Past JEE Main Entrance Paper Set-I

Hint

\(
\begin{aligned}
& \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}- \\
& \left(\frac{1}{2 \times 21}+\frac{1}{4 \times 3}+\ldots+\frac{1}{2024} \cdot \frac{1}{2023}\right)=\frac{1}{2024} \\
& \sum_{r=1}^{1012} \frac{1}{2 r(2 r-1)}=\sum_{r=1}^{1012}\left(\frac{1}{2 r-1}-\frac{1}{2 r}\right) \\
& \quad=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right)
\end{aligned}
\)
\(
\begin{aligned}
& =\left(1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{2023}\right) \\
& -\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2024}\right) \\
& =\left(1+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{1012}\right) \\
& =\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1011}\right) \\
& \frac{-1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1012}\right) \\
& =\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023}-\frac{1}{2024} \\
& \Rightarrow \alpha+1012=2023 \\
& \Rightarrow \alpha=1011
\end{aligned}
\)

Hint

First term is each row form pattern \(2, \quad 5, \quad 11, \quad 20\) and the difference between consecutive numbers forms an A. P. (3, 6, 9)
\(
\begin{aligned}
& \Rightarrow T_n=a n^2+b n+c \\
& \Rightarrow T_1=a+b+c=2 \\
& \Rightarrow T_2=4 a+2 b+c=5 \\
& \Rightarrow T_3=9 a+3 b+c=11 \\
& \Rightarrow 3 a+b=3 \\
& 5 a+b=6 \\
& \Rightarrow 2 a=3 \Rightarrow a=\frac{3}{2}, \quad b=\frac{-3}{2} \Rightarrow c=2 \\
& \Rightarrow T_n=\frac{3}{2} n^2-\frac{3(n)}{2}+2 \Rightarrow \frac{3 n^2-3 n+4}{2} \\
& T_{10}=\frac{3 \times 100-3 \times 10+4}{2}=\frac{274}{2}=137
\end{aligned}
\)
Terms in \(10^{\text {th }}\) row is 10 with 3 differences
\(
\begin{aligned}
& \Rightarrow 137,140,143 \ldots \\
& \Rightarrow \quad S_{10}=\frac{10}{2}(2 \times 137+(10-1) \times 3) \\
& =5(274+27)=5 \times 301=1505
\end{aligned}
\)

Hint

Let 5310 lies in \(k ^{\text {th }}\) row
\(\Rightarrow\) First element of \(k^{\text {th }}\) row is \(\frac{(k-1) k}{2}+1\)
Last element of \(k^{\text {th }}\) row is \(\frac{k(k+1)}{2}\)
\(
\begin{aligned}
& \Rightarrow \frac{(k-1) k}{2}+1 \leq 5310 \leq \frac{k(k+1)}{2} \\
& \Rightarrow k=103
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right)^n C_r \\
& =4 \sum_{r=0}^n r^{2 n} C_r+2 \sum_{r=0}^n r \cdot{ }^n C_r+\sum_{r=0}^n{ }^n C_r \\
& =4 n(n+1) 2^{n-2}+2 \cdot n \cdot 2^{n-1}+2^n \\
& =2^n(n(n+1)+n+1)=2^n(n+1)^2 \\
& \beta=\sum_{r=0}^n\left(\frac{{ }^n C_r}{r+1}\right)+\left(\frac{1}{n+1}\right) \\
& (1+x)^n=\sum_{r=0}^n{ }^n C_r x^r \\
& \int_0^1(1+x)^n d x=\left.\sum_{r=0}^n \frac{{ }^n C_r x^{r+1}}{r+1}\right|_0 ^1=\sum_{r=0} \frac{{ }^n C}{r+1} \\
& \left.\frac{(1+x)^{+1}}{n+1}\right|_0 ^1=\frac{2^n-1}{n+1} \\
& \Rightarrow \beta=\frac{2^{n+1}-1+1}{(n+1)}=\frac{2}{n+1} \\
& \Rightarrow \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n \quad 1)}{\left(\frac{2^{n+1}}{n+1}\right)}=(n+1)^3 \in(140,281) \\
& \Rightarrow(n+1)^3=216 \\
& \Rightarrow n+1=6 \Rightarrow n=5 \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& S(x)=(1+x)+2(1+x)^2+3(1+x)^3+\ldots .+60(1+x)^{60} \\
& (1+x) S(x) \quad=1(1+x)^2+2(1+x)^3+\ldots .+59(1+x)^{60}+60(1+x)^{61} \\
\hline \\
& S(x)[1-1-x]=\left[(1+x)+(1+x)^2+\ldots+(1+x)^{60}\right]-60(1+x)^{61} \\
&
\end{aligned}
\)
\(
\begin{aligned}
& (-x)(S(x))=\frac{(1+x)\left[(1+x)^{60}-1\right]}{(1+x-1)}-60(1+x)^{61} \\
& (-x) S(x)=\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}-60(1+x)^{61} \\
& x S(x)=60(1+x)^{61}-\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}
\end{aligned}
\)
Multiplying \(x\) on both side,
\(
x^2 S(x)=60 x(1+x)^{61}-(1+x)\left[(1+x)^{60}-1\right]
\)
Putting \(x=60\)
\(
\begin{aligned}
& (60)^2 S(60)=60 \times 60(61)^{61}-(61)\left[61^{60}-1\right] \\
& =60 \times 60(61)^{61}-(61) \cdot 61^{60}+61 \\
& =(61)^{61}[60 \times 60-1]+61 \\
& =(3600-1) \cdot 61^{61}+61 \\
& a=3600-1, \quad b=61 \Rightarrow a+b=3660
\end{aligned}
\)

Hint

\(
T_r=3 T_{r-1}+6^r
\)
\(\Rightarrow\) solving homogenous part
\(
T_r=3 T_{r-1}
\)
\(\Rightarrow x=3\) is the root
\(
\therefore T_r=a .3^r
\)
Solving for particular part
\(
\begin{aligned}
& T_r=b .6^r \\
& b .6^r=3 b 6^{r-1}+6^r \\
& \Rightarrow 6 b=3 b+6 \\
& \Rightarrow 3 b=6 \\
& \Rightarrow b=2 \\
& T_r=a^n+a^p \\
& T_r=a 3^{b r}+2.6^r \dots(i) \\
& T_r=3 T_{r-1}+6^r
\end{aligned}
\)
Putting \(r=2\)
\(
T_2=18+36=54 \dots(ii)
\)
Using equation (i) and (ii)
\(
\begin{aligned}
& 54=9 a+72 \Rightarrow-18=9 a \Rightarrow a=-2 \\
& \therefore T_r=2 \cdot 6^r-2 \cdot 3^r=2\left(6^r-3^r\right) \\
& \sum_{r=1}^n T_r=2 \sum 6^r-2 \sum 3^r \\
& =2 \cdot 6 \frac{\left(6^n-1\right)}{5}-2 \cdot 3 \frac{\left(3^n-1\right)}{2} \\
& =\frac{3}{5}\left(4 \cdot 6^n-5 \cdot 3^n+1\right) \\
& \therefore n^2-12 n+39=3 \\
& n^2-12 n+36=0 \\
& (n-6)^2=0 \\
& \therefore n=6
\end{aligned}
\)

Hint

\(
\begin{aligned}
& S=1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\ldots \infty \\
& =1+\frac{(1-\sqrt{2} / \sqrt{3})}{2}+\frac{(1-\sqrt{2} / \sqrt{3})^2}{6}+\frac{(1-\sqrt{2} / \sqrt{3})^3}{12}+\ldots \infty
\end{aligned}
\)
let \(1-\frac{\sqrt{2}}{\sqrt{3}}=a\)
\(
\begin{aligned}
& S=1+\frac{a}{2}+\frac{a^2}{6}+\frac{a^3}{12}+\ldots \\
& =1+\left(1-\frac{1}{2}\right) a+\left(\frac{1}{2}-\frac{1}{3}\right) a^2+\left(\frac{1}{3}-\frac{1}{4}\right) a^3+\ldots \\
& =1+\left(a+\frac{a^2}{2}+\frac{a^3}{3} \ldots \infty\right)+\frac{1}{a}\left(\frac{-a^2}{2}-\frac{a^3}{3}-\frac{a^4}{4} \ldots \infty\right) \\
& =-\ln (1-a)+\frac{1}{a}\left(-a-\frac{a^2}{2}-\frac{a^3}{3} \ldots \infty\right)+2 \\
& =-\ln (1-a)+\frac{1}{a} \ln (1-a)+2 \\
& =2+\left(\frac{1}{a}-1\right) \ln (1-a)
\end{aligned}
\)
\(
\begin{aligned}
& =2+\left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1\right) \ln \left(1-1+\sqrt{\frac{2}{3}}\right) \\
& =2+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2} \ln \sqrt{\frac{2}{3}} \\
& =2+\left(\frac{\sqrt{6}+2}{1} \cdot \frac{1}{2} \ln \frac{2}{3}\right) \\
& \therefore 2+\left(\sqrt{\frac{3}{2}}+1\right) \ln \frac{2}{3} \\
& \therefore 11 a+18 b=76
\end{aligned}
\)

Hint

Let \(a_n=a+(n-1) d \forall n \in N\)
\(
\begin{aligned}
& A_k=\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\ldots a_{2 k-1}^2-a_{2 k}^2 \\
&=(-d)\left(a_1+a_2+\ldots+a_{2 k}\right) \\
& A_k=(-d k)(2 a+(2 k-1) d) \\
& \Rightarrow A_3=(-3 d)(2 a+5 d)=-153 \\
& \Rightarrow d(2 a+5 d)=51 \quad \ldots \text { (i) } \\
& A_5=(-5 d)(2 a+9 d)=-435 \\
& \Rightarrow d(2 a+9 d)=87 \\
& \Rightarrow 4 d^2=36 \Rightarrow d= \pm 3(d=3 \text { positive terms }) \\
& \Rightarrow 3(2 a+27)=87 \\
& \Rightarrow 2 a=29-27 \\
& \Rightarrow a=1 \\
& a_{17}-A_7=(a+16 d)-(-7 d)(2+13 d) \\
&=49+7 \times 3(2+39) \\
&=49+21 \times 41=910
\end{aligned}
\)

Hint

To solve this problem, let’s first denote the three successive terms of a geometric progression (G.P.) with common ratio \(r\) as \(a, a r\), and \(a r^2\), where \(a\) is the first term and \(r>1\). These three terms represent the lengths of the sides of a triangle.
According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, for the three terms to form a triangle, the following inequalities must hold:
1) \(a+a r>a r^2\)
2) \(a+a r^2>a r\)
3) \(a r+a r^2>a\)
Given that \(r>1\), inequalities 2 and 3 will always hold because:
\(
a r<a r^2 \text { and } a<a r
\)
indicating that both \(a+a r^2\) and \(a r+a r^2\) will be greater than \(a r\) and \(a\) respectively. Therefore, we only need to check the first inequality to ensure that the three terms can form a triangle:
\(
a+a r>a r^2
\)
Simplifying the right side by factoring \(r\) :
\(
1>r(r-1)
\)
Given that \(r>1\), the quantity \((r-1)\) is positive; hence, \(r(r-1)\) is also positive. This means the actual value for \(r\) to satisfy the inequality is within the interval \((1, \sqrt{2})\) because \(r(r-1)\) increases with increasing \(r\), and it would be 1 when \(r=\sqrt{2}\). It should be greater than 1 , and less than \(\sqrt{2}\) such that \(r^2-r\) stays below 1 .
Now let’s consider the expressions \([r]\) and \([-r]\). The symbol \([x]\) denotes the greatest integer less than or equal to \(x\) (also known as the floor function).
Since \(1<r<\sqrt{2},[r]=1\), because 1 is the greatest integer less than \(r\) within that interval.
For \([-r]\), we need the greatest integer less than or equal to \(-r\). Since \(-r\) is negative and less than -1 (because \(r>1\) ), \([-r]=-2\), as this is the greatest integer that does not exceed the negative value of \(r\) (which lies between \(-\sqrt{2}\) and -1 ).
Now we can substitute these values into the expression:
\(
3[r]+[-r]=3 \cdot 1+(-2)=3-2=1
\)
Therefore \(3[r]+[-r]\) is equal to 1 .

Hint

To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).
The first AP is:
\(
3,7,11,15, \ldots, 403
\)
The common difference \(\left(d_1\right)\) for the first AP can be calculated by subtracting the first term from the second term:
\(
d_1=7-3=4
\)
The second AP is:
\(
2,5,8,11, \ldots, 404
\)
The common difference \(\left(d_2\right)\) for the second AP is:
\(
d_2=5-2=3
\)
To find the terms common to both sequences, we need to find a term that appears in both sequences. Any common term must be of the form \(3+4 k\) and \(2+3 l\) for some integers \(k\) and \(l\). We want to find when these two forms will give us the same number, so we set them equal to each other:
\(
3+4 k=2+3 l
\)
Rearranging the terms gives us:
\(
4 k-3 l=2-3
\)
This simplifies to:
\(
4 k-3 l=-1 \ldots \ldots(1)
\)
The solutions to equation (1) will give us the common terms. Notice this is a Diophantine equation (A Diophantine equation is a polynomial equation, usually with two or more variables,) and has an infinite number of solutions. Let’s find one such solution. We can see that:
\(
k=1 \text { yields } 4(1)-3 l=-1 \Longrightarrow 4-3 l=-1 \Longrightarrow 3 l=5 \Longrightarrow l=
\)
This is not an integer solution for \(l\), so \(k=1\) does not work. Trying \(k=2\) gives:
\(
4(2)-3 l=-1 \Longrightarrow 8-3 l=-1 \Longrightarrow 3 l=9 \Longrightarrow l=3
\)
Now we’ve found integers \(k=2\) and \(l=3\) that satisfy the equation. The corresponding term in both sequences would be:
\(
3+4(2)=3+8=11 \text { and } 2+3(3)=2+9=11
\)
Since 11 is a common term, we can assert that every common term in both APs will be of the form \(11+m(4 \times 3)\), where \(m\) is a non-negative integer, and \(4 \times 3=12\) is the LCM of the common differences of the two APs. Thus, the general form for the common terms would be:
\(
11+12 m
\)
Now we are to find all terms that are common up to 403 in the first sequence and up to 404 in the second sequence. Because the first sequence doesn’t exceed 403 , we’ll use this as our limit:
\(
11+12 m \leq 403
\)
To find the largest possible integer value for \(m\), we solve the inequality:
\(
\begin{aligned}
& 12 m \leq 403-11 \\
& 12 m \leq 392 \\
& m \leq 32 \frac{2}{3}
\end{aligned}
\)
Since \(m\) has to be an integer, the largest possible value for \(m\) is 32 . Therefore, the common terms are generated by \(m=0,1,2, \ldots, 32\). There are \(32+1=33\) terms in total.
We will now sum these up. The sum of an AP is given by the formula:
\(
S=\frac{n}{2}\left(a_1+a_n\right)
\)
Where \(S\) is the sum, \(n\) is the number of terms, \(a_1\) is the first term, and \(a_n\) is the last term. Using the formula:
\(
\begin{aligned}
& S=\frac{33}{2}(11+(11+12 \times 32)) \\
& S=\frac{33}{2}(11+11+384) \\
& S=\frac{33}{2}(11+11+384)
\end{aligned}
\)
\(
\begin{aligned}
& S=\frac{33}{2}(406) \\
& S=33 \times 203 \\
& S=6699
\end{aligned}
\)
Therefore, the sum of the common terms in the two arithmetic progressions is 6699.

Hint

\(
\begin{aligned}
& S _{ n }=3+7+11+\ldots \ldots n \text { terms } \\
& =\frac{ n }{2}(6+( n -1) 4)=3 n +2 n ^2-2 n \\
& =2 n ^2+ n \\
& \sum_{ k =1}^{ n } S _{ k }=2 \sum_{ k =1}^{ n } K ^2+\sum_{ k =1}^{ n } K \\
& =2 \cdot \frac{ n ( n +1)(2 n +1)}{6}+\frac{ n ( n +1)}{2} \\
& = n ( n +1)\left[\frac{2 n +1}{3}+\frac{1}{2}\right] \\
& =\frac{ n ( n +1)(4 n +5)}{6} \\
& \Rightarrow 40<\frac{6}{ n ( n +1)} \sum_{k=1}^n S_{ k }<42 \\
& 40<4 n +5<42 \\
& 35<4 n<37 \\
& n =9
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \alpha=1^2+4^2+8^2 \ldots \\
& \quad t_n=a^2+b n+c \\
& 1=a+b+c \\
& 4=4 a+2 b+c \\
& 8=9 a+3 b+c
\end{aligned}
\)
On solving we get, \(a =\frac{1}{2}, b =\frac{3}{2}, c =-1\)
\(
\begin{aligned}
& \alpha=\sum_{n=1}^{10}\left(\frac{n^2}{2}+\frac{3 n}{2}-1\right)^2 \\
& 4 \alpha=\sum_{n=1}^{10}\left(n^2+3 n-2\right)^2, \beta=\sum_{n=1}^{10} n^4 \\
& 4 \alpha-\beta=\sum_{n=1}^{10}\left(6 n^3+5 n^2-12 n+4\right)=55(353)+40
\end{aligned}
\)

Hint

\(
8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}
\)
(sum of infinite terms of A.G.P \(=\frac{a}{1-r}+\frac{d r}{(1-r)^2}\) )
\(
\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9
\)

Hint

We can rewrite the given series as follows :
\(
S=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right)+\left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right)+\left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right)+\ldots
\)
The first few terms of the series are :
\(
S=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\ldots\right)+\left(\frac{1}{9}+\frac{1}{18}+\frac{1}{36}+\ldots\right)-\left(\frac{1}{27}+\frac{1}{54}+\ldots\right)+\ldots
\)
We can now see that each group of terms forms a geometric series with a common ratio of \(\frac{1}{2}\) :
\(
S=\frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{\frac{1}{3}}{1-\frac{1}{2}}+\frac{\frac{1}{9}}{1-\frac{1}{2}}-\frac{\frac{1}{27}}{1-\frac{1}{2}}+\ldots
\)
The series can be rewritten as :
\(
S=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\ldots\right)
\)
Now, we can simplify and rewrite the series inside the parentheses as:
\(
S=2\left[\frac{1}{2}+\left(-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\ldots\right)\right]
\)
The series inside the parentheses is an infinite geometric series with the first term \(a=-\frac{1}{3}\) and the common ratio \(r=-\frac{1}{3}\) :
\(
\begin{aligned}
& S=2\left(\frac{1}{2}+\frac{a}{1-r}\right)=2\left(\frac{1}{2}+\frac{-\frac{1}{3}}{1+\frac{1}{3}}\right)=2\left(\frac{1}{2}+\frac{-\frac{1}{3}}{\frac{4}{3}}\right) \\
& =2\left(\frac{1}{2}-\frac{1}{3} \times \frac{3}{4}\right) \\
& =2\left(\frac{1}{2}-\frac{1}{4}\right)=2\left(\frac{1}{4}\right)=\frac{1}{2}
\end{aligned}
\)
Thus, the sum of the series is \(\frac{1}{2}\), and \(\alpha=1\) and \(\beta=2\) are co-prime.
Therefore, \(\alpha+3 \beta=1+6=7\)

Hint

\(
\begin{aligned}
& \sum_{r=1}^{10}\left(2 \cdot(2 r)^2-(2 r+1)^2\right) \\
& =\sum_{r=1}^{10}\left(8 r^2-4 r^2-4 r-1\right) \\
& =\sum_{r=1}^{10}\left(4 r^2-4 r-1\right) \\
& =\frac{4 \cdot 10 \cdot 11 \cdot 21}{6}-4 \frac{10 \cdot 11}{2}-10 \\
& =44 \cdot 35-220-10 \\
& =1540-230=1310
\end{aligned}
\)

Hint

From the given series :
\(
10=1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots
\)
We isolate the 1 to get :
\(
9=\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots(1)
\)
Divide each term in the equation by \(k\) :
\(
\frac{9}{k}=\frac{4}{k^2}+\frac{8}{k^3}+\frac{13}{k^4}+\ldots(2)
\)
Subtracting the second equation from the first, we obtain :
\(
9\left(1-\frac{1}{k}\right)=\frac{4}{k}+\frac{4}{k^2}+\frac{5}{k^3}+\frac{6}{k^4}+\ldots
\)
This is equivalent to :
\(
S=9\left(1-\frac{1}{k}\right)=\frac{4}{k}+\frac{4}{k^2}+\frac{5}{k^3}+\frac{6}{k^4}+\ldots(3)
\)
Divide both sides by \(k\) again, we get : \(\frac{S}{k}=\frac{4}{k^2}+\frac{4}{k^3}+\frac{5}{k^4}+\ldots(4)\)
Subtracting equation (4) from (3), we have :
\(
\left(1-\frac{1}{k}\right) S=\frac{4}{k}+\frac{1}{k^3}+\frac{1}{k^4}+\ldots
\)
In this equation, you’ve treated the infinite series on the right-hand side as a geometric series with a ratio of \(1 / k\), so the sum of this series can be expressed as \(\frac{1 / k^3}{1-1 / k}\).
Therefore, we get :
\(
9\left(1-\frac{1}{k}\right)^2=\frac{4}{k}+\frac{1 / k^3}{1-1 / k}
\)
Now, simplifying this equation, we get :
\(
\begin{aligned}
& 9(k-1)^2=4 k^2-4 k+1 \\
& \Rightarrow 9\left(k^2-2 k+1\right)=4 k^2-4 k+1 \\
& \Rightarrow 9 k^2-18 k+9=4 k^2-4 k+1 \\
& \Rightarrow 9 k^2-4 k^2-18 k+4 k+9-1=0 \\
& \Rightarrow 5 k^2-14 k+8=0
\end{aligned}
\)
This is a standard form quadratic equation, \(a x^2+b x+c=0\), where \(a=5, b=-14\), and \(c=8\).
The solutions can be found using the quadratic formula, \(k=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\).
Substituting the values for \(a, b\), and \(c\), we have :
\(
\begin{aligned}
& \Rightarrow k=\frac{14 \pm \sqrt{(-14)^2-4 \times 5 \times 8}}{2 \times 5} \\
& \Rightarrow k=\frac{14 \pm \sqrt{196-160}}{10} \\
& \Rightarrow k=\frac{14 \pm \sqrt{36}}{10} \\
& \Rightarrow k=\frac{14 \pm 6}{10}
\end{aligned}
\)
So, the solutions are \(k=2\) or \(k=0.8\). However, since the question mentions \(k \in N\), i.e., \(k\) is a natural number, the only valid solution is \(k=2\).

Hint

We have, \(S=109+\frac{108}{5}+\frac{107}{5^2}+\ldots+\frac{2}{5^{107}}+\frac{1}{5^{108}} \dots(i)\)
\(\frac{S}{5}=\frac{109}{5}+\frac{108}{5^2}+\frac{107}{5^3}+\ldots+\frac{2}{5^{108}}+\frac{1}{5^{109}} \dots(ii)\)
On subtracting Eq. (ii) from Eq. (i), we get
\(
\begin{aligned}
& \frac{4 S}{5}=109-\frac{1}{5}-\frac{1}{5^2}-\ldots .-\frac{1}{5^{108}}-\frac{1}{5^{109}} \\
& \frac{4 S}{5}=109-\left[\frac{1}{5}+\frac{1}{5^2}+\ldots+\frac{1}{5^{108}}+\frac{1}{5^{109}}\right]
\end{aligned}
\)
This form a GP with \(r=\frac{1}{5}\)
\(
\begin{aligned}
& \frac{4 S}{5}=109-\frac{1}{5}\left[\frac{1-\frac{1}{5^{109}}}{1-\frac{1}{5}}\right] \\
& =109-\frac{1}{4}\left[1-\frac{1}{5^{109}}\right] \\
& =109-\frac{1}{4}+\frac{1}{4} \times \frac{1}{5^{109}} \\
& \therefore 4 S=\frac{5}{4}\left[435+\frac{1}{5^{1009}}\right] \\
& \Rightarrow 16 S=2175+\frac{1}{5^{108}} \\
& 16 S-(25)^{-54}=2175 \\
&
\end{aligned}
\)

Hint

Since, common ratio of A.G.P. is 2 therefore A.G.P. can be taken as
\(
\begin{aligned}
& \frac{(a-2 d)}{4}, \frac{(c-d)}{2}, a, 2(a+d), 4(a+2 d) \\
& \text { or } a_1, a_2, 2, a_3, a_4 \text { (Given) } \\
& \Rightarrow a=2
\end{aligned}
\)
also sum of thes A.G.P. is \(\frac{49}{2}\)
\(
\begin{aligned}
& \Rightarrow \frac{2-2 d}{4}+\frac{2-d}{2}+2+2(2+d)+4(2+2 d)=\frac{49}{2} \\
& \Rightarrow \frac{1}{4}[2-2 d+4-2 d+8+16+8 d+32+32 d]=\frac{49}{2} \\
& \Rightarrow 36 d+62=98 \\
& \Rightarrow 36 d=36 \Rightarrow d=1
\end{aligned}
\)
Hence, \(a_4=4(a+2 d)=4(2+2 \times 1)=16\)

Hint

\(\because \frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in A.P.
\(\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{2}{y} \dots(i)\)
and \(x, \sqrt{2} y, z\) are in G.P.
\(\Rightarrow 2 y^2=x z \dots(ii)\)
\(
\begin{aligned}
& \text { from (i), } \frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2} \\
& \Rightarrow 4 y=x+z \\
& \text { Also, } x y+y z+z x=\frac{3}{\sqrt{2}} x y z \\
& y(4 y)+x z=\frac{3}{\sqrt{2}}\left(2 y^2\right) y \\
& \Rightarrow 4 y^2+2 y^2=3 \sqrt{2} y^3 \\
& \Rightarrow 6 y^2=3 \sqrt{2} y^3 \Rightarrow y=\sqrt{2} \\
& \therefore 3(x+y+z)^2=3(5 y)^2=3(5 \sqrt{2})^2 \\
& =150
\end{aligned}
\)

Hint

\(
(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17}+\ldots \ldots+20(21)^{19}=k(20)^{19}
\)
\(
\Rightarrow(20)^{19}\left[1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19}\right]=k(20)^{19}
\)
\(
\Rightarrow k=1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19} \dots(i)
\)
Now,
\(
k\left(\frac{21}{20}\right)=\left(\frac{21}{20}\right)+2\left(\frac{21}{20}\right)^2+3\left(\frac{21}{20}\right)^3+\ldots+20\left(\frac{21}{20}\right)^{20} \dots(ii)
\)
On subtracting Equation (ii) from Equation (i), we get
\(
\begin{aligned}
& k\left(\frac{-1}{20}\right)=1+\frac{21}{20}+\left(\frac{21}{20}\right)^2+\ldots \ldots+\left(\frac{21}{20}\right)^{19}-20\left(\frac{21}{20}\right)^{20} \\
& \Rightarrow k\left(\frac{-1}{20}\right)=\frac{1\left(\left(\frac{21}{20}\right)^{20}-1\right)}{\frac{21}{20}-1}-20\left(\frac{21}{20}\right)^{20} \\
& \Rightarrow k\left(\frac{-1}{20}\right)=20\left(\left(\frac{21}{20}\right)^{20}-1\right)-20\left(\frac{21}{20}\right)^{20} \\
& =20\left(\frac{21}{20}\right)^{20}-20-20\left(\frac{21}{20}\right)^{20} \\
& \Rightarrow k\left(\frac{-1}{20}\right)=-20 \\
& \Rightarrow k=400
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
3,7,11,15, \ldots \ldots \ldots \ldots 399: & d _1=4 \\
2,5,8,11, \ldots \ldots \ldots \ldots, 359: & d _2=3 \\
2,7,12,17, \ldots \ldots, 197: & d _3=5
\end{array}
\)
\(
\operatorname{LCM}\left( d _1, d _2, d _3\right)=60
\)
Common terms are \(47,107,167\)
\(
\text { Sum }=321
\)

Hint

\(
\begin{aligned}
& a_1+a_2+a_3+a_4=50 \\
& \Rightarrow 32+6 d=50 \\
& \Rightarrow d=3 \\
& \text { and, } a_{n-3}+a_{n-2}+a_{n-1}+a_n=170 \\
& \Rightarrow 32+(4 n-10) \cdot 3=170 \\
& \Rightarrow n =14 \\
& a_7=26, a_8=29 \\
& \Rightarrow a_7 \cdot a_8=754
\end{aligned}
\)

Hint

\(
S=1^2-2.3^2+3.5^2-4.7^2+\ldots \ldots+15.29^2
\)
Separating odd placed and even placed terms we get
\(
\begin{aligned}
& S =\left(1.1^2+3.5^2+\ldots .15 \cdot(29)^2\right)-\left(2.3^2+4.7^2\right. \\
& +\ldots+14 \cdot(27)^2 \\
& =\sum_{r=1}^8(2 r-1)(4 r-3)^2-\sum_{r=1}^7 2 r(4 r-1)^2 \\
& =\sum_{r=1}^8\left(32 r^3-64 r^2+42 r-9\right)-2 \sum_{r=1}^7 16 r^3-8 r^2+r \\
& =32 \times 36^2-64 \times 204+1512-72 \\
& -2\left(16 \times 28^2-1120+28\right) \\
& =6592
\end{aligned}
\)

Hint

\(
\begin{aligned}
& a_{11}=18 \\
& a+10 d=18 \dots(i) \\
& a_5=2 a_7 \\
& a+4 d=2(a+6 d) \\
& a=-8 d \dots(ii)
\end{aligned}
\)
\(
\text { (i) and (ii) } \Rightarrow a=-72, d=9 \text {. }
\)
On rationalising the denominator, given expression
\(
\begin{aligned}
& =12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{11}}}{-d}+\frac{\sqrt{a_{11}}-\sqrt{a_{12}}}{-d}+\ldots+\frac{\sqrt{a_{17}}-\sqrt{a_{18}}}{-d}\right] \\
& =12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{18}}}{-d}\right] \\
& =12\left[\frac{\sqrt{a_{11}-d}-\sqrt{a_{11}+7 d}}{-d}\right] \\
& =12\left[\frac{\sqrt{18-9}-\sqrt{18+63}}{-9}\right]=12 \times \frac{2}{3}=8
\end{aligned}
\)

Hint

First common term is 11
Common difference of series of common terms is \(\operatorname{LCM}(4,5)=20\)
\(
\begin{aligned}
& a_8=a+7 d \\
& =11+7 \times 20=151
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sum_{n=0}^{\infty} \frac{n^3(2 n!)+(2 n-1)(n!)}{n!\cdot(2 n)!} \\
& =\sum_{n=0}^{\infty} \frac{n^3}{n!}+\frac{2 n-1}{2 n!} \\
& =\sum_{n=0}^{\infty} \frac{3}{(n-2)!}+\frac{1}{(n-3)!}+\frac{1}{(n-1)!}+\frac{1}{(2 n-1)!}-\frac{1}{(2 n)!} \\
& =3 e+e+e-\frac{1}{e} \\
& =5 e-\frac{1}{e} \\
& \therefore a=5, b=-1, c=0 \\
& \therefore a^2-b+c=26
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \because a_n=a_{n-1}+(n-1) \text { and } a_1=b_1=1 \\
& b_n=b_{n-1}+a_{n-1} \\
& \therefore b_{n+1}=2 b_n-b_{n-1}+n-1
\end{aligned}
\)
\(
\begin{array}{|c|c|c|}
\hline n & b_n & b_n-n \\
\hline 1 & 1 & 0 \\
\hline 2 & 2 & 0 \\
\hline 3 & 4 & 1 \\
\hline 4 & 8 & 4 \\
\hline 5 & 15 & 10 \\
\hline 6 & 26 & 20 \\
\hline 7 & 42 & 35 \\
\hline 8 & 64 & 56 \\
\hline 9 & 93 & 84 \\
\hline 10 & 130 & 120 \\
\hline
\end{array}
\)
\(
\begin{aligned}
& \therefore 2 S-T=\left(\sum_{n=1}^8 \frac{b_n-n}{2^{n-1}}\right)+\frac{b_9}{2^8}+\frac{b_{10}}{2^9} \\
& =\frac{461}{128} \\
& \therefore 2^7(2 S-T)=461
\end{aligned}
\)

Hint

\(\left\{a_k\right\}\) be a G.P. with \(a_1=4, r=r_1\)
and
\(\left\{b_k\right\}\) be G.P. with \(b_1=4, r=r_2\left(r_1<r_2\right)\)
Now
\(
\begin{aligned}
& C_k=a_k+b_k \\
& c_1=4+4=8 \text { and } c_2=5 \\
& a_2+b_2=5 \\
& \therefore r_1+r_2=\frac{5}{4} \\
& \text { and } c_3=\frac{13}{4} \Rightarrow r_4^2+r_2^2=\frac{13}{16} \\
& \therefore \frac{25}{16}-2 r_1 r_2=\frac{13}{16} \Rightarrow 2 r_1 r_2=\frac{3}{4} \\
& \therefore r_2-r_1=\sqrt{\frac{25}{16}-\frac{3}{2}}=\frac{1}{4} \\
& \therefore r_2=\frac{3}{4}, r_1=\frac{1}{2} \\
& \therefore a_6=4 \times \frac{1}{2^5}=\frac{1}{8}, b_4=4 \times \frac{27}{64}=\frac{27}{16} \\
& \text { and } \sum_{K=1}^{\infty} C_K=4\left[\frac{1}{1-\frac{1}{2}}+\frac{1}{1-\frac{3}{4}}\right]=24 \\
& \therefore \sum_{K=1}^{\infty} C_K-\left(12 a_6+8 b_4\right)=09
\end{aligned}
\)

Hint

Let \(r\) be the common ratio of the G.P
\(
\begin{aligned}
& \therefore a_1 r^3 \times a_1 r^5=9 \\
& a_1^2 r^8=9 \Rightarrow a_1 r^4=3
\end{aligned}
\)
and
\(
\begin{gathered}
a_1\left(r^4+r^6\right)=24 \\
\Rightarrow 3\left(1+r^2\right)=24 \\
\therefore r^2=7 \text { and } a_1=\frac{3}{49}
\end{gathered}
\)
Now
\(
\begin{aligned}
& a_1 a_9+a_2 a_4 a_9+a_5+a_7 \\
& =a_1^2 r^8+a_1^3 r^{12}+24 \\
& =24+\frac{9}{7^4} \times 7^4+\frac{27}{7^6} \cdot 7^6=60
\end{aligned}
\)

Hint

\(
\begin{aligned}
& a , b , \frac{1}{18} \rightarrow GP \\
& \frac{ a }{18}= b ^2 \quad \ldots(i)
\end{aligned}
\)
\(
\begin{aligned}
& \frac{1}{ a }, 10, \frac{1}{ b } \rightarrow AP \\
& \frac{1}{ a }+\frac{1}{ b }=20
\end{aligned}
\)
\(\Rightarrow a + b =20 ab\), from eq. (i) ; we get
\(
\begin{aligned}
& \Rightarrow 18 b ^2+ b =360 b ^3 \\
& \Rightarrow 360 b ^2-18 b -1=0 \quad\{\because b \neq 0\} \\
& \Rightarrow b =\frac{18 \pm \sqrt{324+1440}}{720} \\
& \Rightarrow b=\frac{18+\sqrt{1764}}{720} \quad\{\because b >0\} \\
& \Rightarrow b=\frac{1}{12} \\
& \Rightarrow a =18 \times \frac{1}{144}=\frac{1}{8}
\end{aligned}
\)
Now, \(16 a+12 b=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3\)

Hint

Given \(\frac{1^3+2^3+3^3+\ldots \text { up to } \quad n \text { terms }}{1.3+2.5+3.7+\ldots \text { up to } \quad n \text { terms }}=\frac{9}{5}\)
Now, Let \(S=1.3+2.5+3.7+\ldots\)
\(T_n=n \cdot(2 n+1)\)
\(\therefore S=\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\)
\(\Rightarrow \frac{\left(\frac{n(n+1)}{2}\right)^2}{n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]}=\frac{9}{5}\)
\(\Rightarrow 5 n^2-19 n-30=0\)
\(\Rightarrow(5 n+6)(n-5)=0\)
\(\therefore n=5\)

Hint

\(
\begin{aligned}
& T_4=500 \\
& a r^3=500 \Rightarrow a=\frac{500}{r^3}
\end{aligned}
\)
Now,
\(
\begin{aligned}
& S_6>S_5+1 \\
& \frac{a\left(1-r^6\right)}{1-r}-\frac{a\left(1-r^5\right)}{1-r}>1 \\
& a r^5>1
\end{aligned}
\)
Now, \(r=\frac{1}{m}\) and \(a=\frac{500}{r^3}\)
\(
\begin{aligned}
& \Rightarrow \quad m^2<500 \\
& \because m>0 \Rightarrow m \in(0,10 \sqrt{5}) \\
& S_7<S_6+\frac{1}{2} \\
& \frac{a\left(1-r^6\right)}{1-r}<\frac{a\left(1-r^6\right)}{1-r}+\frac{1}{2} \\
& a r^6<\frac{1}{2} \\
& \because r=\frac{1}{m} \text { and } a=\frac{500}{r^5} \\
& \frac{1}{m^3}<\frac{1}{1000} \\
& \Rightarrow m \in(10, \infty) \\
&
\end{aligned}
\)
Possible values of \(m\) is \(\{11,12, \ldots .22\}\)
\(
\because m \in N
\)
Total 12 values

Hint

Given
\(
\begin{aligned}
& S=\frac{a_1}{2}+\frac{a_2}{2^2}+\frac{a_3}{2^3}+\frac{a_4}{2^4}+\ldots \ldots \infty \\
& \frac{\frac{1}{2} S=\frac{a_1}{2^2}+\frac{a_2}{2^3}+\ldots \ldots \ldots \infty}{\frac{S}{2}=\frac{a_1}{2}+\frac{\left(a_2+a_1\right)}{2^2}+\frac{\left(a_3+a_2\right)}{2^3}+\ldots \ldots \infty} \\
& \Rightarrow \frac{S}{2}=\frac{a_1}{2}+\frac{d}{2} \\
& \Rightarrow a_1+d=a_2=4 \Rightarrow 4 a_2=16
\end{aligned}
\)

Hint

\(
\begin{aligned}
& S=\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\frac{1}{100 \times 101 \times 102} \\
& =\frac{1}{(3-1) .1}\left[\frac{1}{2 \times 3}-\frac{1}{101 \times 102}\right] \\
& =\frac{1}{2}\left(\frac{1}{6}-\frac{1}{101 \times 102}\right) \\
& =\frac{143}{102 \times 101}=\frac{k}{101} \\
& \therefore 34 k=286
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{1}{3^{12}}+5\left(\frac{2^0}{3^{12}}+\frac{2^1}{3^{11}}+\frac{2^2}{3^{10}}+\ldots \ldots+\frac{2^{11}}{3}\right)=2^n \cdot m \\
& \Rightarrow \frac{1}{3^{12}}+5\left(\frac{1}{3^{12}} \frac{\left((6)^2-1\right)}{(6-1)}\right)=2^n \cdot m \\
& \Rightarrow \frac{1}{3^{12}}+\frac{5}{5}\left(\frac{1}{3^{12}} \cdot 2^{12} \cdot 3^{12}-\frac{1}{3^{12}}\right)=2^n \cdot m \\
& \Rightarrow \frac{1}{3^{12}}+2^{12}-\frac{1}{3^{12}}=2^n \cdot m \\
& \Rightarrow 2^n \cdot m=2^{12} \\
& \Rightarrow m=1 \text { and } n=12 \\
& m \cdot n=12
\end{aligned}
\)

Hint

\(
\begin{aligned}
& T_n=\frac{\sum_{k=1}^n\left[(2 k)^3-(2 k-1)^3\right]}{n(4 n+3)} \\
& =\frac{\sum_{k=1}^n 4 k^2+(2 k-1)^2+2 k(2 k-1)}{n(4 n+3)} \\
& =\frac{\sum_{k=1}^n\left(12 k^2-6 k+1\right)}{n(4 n+3)} \\
& =\frac{2 n\left(2 n^2+3 n+1\right)-3 n^2-3 n+n}{n(4 n+3)} \\
& =\frac{n^2(4 n+3)}{n(4 n+3)}=n \\
& \therefore T_n=n \\
& S_n=\sum_{n=1}^{15} T_n=\frac{15 \times 16}{2}=120
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sum_{k=1}^{10} \frac{k}{k^4+k^2+1} \\
& =\frac{1}{2}\left[\sum_{k=1}^{10}\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)\right. \\
& =\frac{1}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\ldots+\frac{1}{91}-\frac{1}{111}\right] \\
& =\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{110}{2.111}=\frac{55}{111}=\frac{m}{n} \\
& \therefore m+n=55+111=166
\end{aligned}
\)

Hint

\(
\begin{aligned}
& d_1=\frac{199-100}{2} \notin I \\
& d_2=\frac{199-100}{3}=33 \\
& d_3=\frac{199-100}{4} \notin I \\
& d_n=\frac{199-100}{i+1} \in I \\
& d_i=33+11,9 \\
& \text { Sum of CD’s }=33+11+9 \\
& =53
\end{aligned}
\)

Hint

Given series
\(
\{3 \times 1\},\{3 \times 2,3 \times 3,3 \times 4\},\{3 \times 5,3 \times 6,3 \times 7,3 \times 8,3 \times 9\}, \ldots .
\)
\(
[latex]\therefore 11^{\text {th }}\) set will have \(1+(10) 2=21\) term
Also upto \(10^{\text {th }}\) set total \(3 \times k\) type terms will be
\(
1+3+5+\ldots \ldots+19=100-\text { term }
\)
\(
\begin{aligned}
& \therefore \text { Set } 11=\{3 \times 101,3 \times 102, \ldots \ldots 3 \times 121\} \\
& \therefore \text { Sum of elements }=3 \times(101+102+\ldots+121) \\
& =\frac{3 \times 222 \times 21}{2}=6993
\end{aligned}
\)

Hint

\(\because\) Roots of \(2 a x^2-8 a x+1=0\) are \(\frac{1}{p}\) and \(\frac{1}{r}\) and roots of \(6 b x^2+12 b x+1=0\) are \(\frac{1}{q}\) and \(\frac{1}{s}\).
Let \(\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}\) as \(\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta\)
So sum of roots \(2 \alpha-2 \beta=4\) and \(2 \alpha+2 \beta=-2\)
Clearly \(\alpha=\frac{1}{2}\) and \(\beta=-\frac{3}{2}\)
Now product of roots, \(\frac{1}{p} \cdot \frac{1}{r}=\frac{1}{2 a}=-5 \Rightarrow \frac{1}{a}=-10\) and \(\frac{1}{q} \cdot \frac{1}{x}=\frac{1}{6 b}=-8 \Rightarrow \frac{1}{b}=-48\)
So, \(\frac{1}{a}-\frac{1}{b}=38\)

Hint

Given,
\(
\begin{aligned}
& a_n=a_{n-1}+2 \\
& \Rightarrow a_n-a_{n-1}=2
\end{aligned}
\)
\(\therefore\) In this series between any two consecutives terms difference is 2 . So this is an A.P. with common difference 2 .
Also given \(a_1=1\)
\(\therefore\) Series is \(=1,3,5,7 \ldots\)
\(
\therefore a_n=1+(n-1) 2=2 n-1
\)
Also \(b_n=a_n+b_{n-1}\)
When \(n=2\) then
\(
\begin{aligned}
& b_2-b_1=a_2=3 \\
& \Rightarrow b_2-1=3 \text { [Given } b_1=1 \text { ] } \\
& \Rightarrow b_2=4
\end{aligned}
\)
When \(n=3\) then
\(
\begin{aligned}
& b_3-b_2=a_3 \\
& \Rightarrow b_3-4=5 \\
& \Rightarrow b_3=9
\end{aligned}
\)
\(\therefore\) Series is \(=1,4,9 \ldots\)
\(=1^2, 2^2, 3^2 \ldots\) \(n^2\)
\(
\therefore b_n=n^2
\)
\(
\begin{aligned}
& \text { Now, } \sum_{n=1}^{15}\left(a_n \cdot b_n\right) \\
& =\sum_{n=1}^{15}\left[(2 n-1) n^2\right] \\
& =\sum_{n=1}^{15} 2 n^3-\sum_{n=1}^{15} n^2 \\
& =2\left(1^3+2^3+\ldots 15^3\right)-\left(1^2+2^2+\ldots 15^2\right) \\
& =2 \times\left(\frac{15 \times 16}{2}\right)^2-\left(\frac{15(16) \times 31}{6}\right) \\
& =27560
\end{aligned}
\)

Hint

Given,
\(
\begin{aligned}
& f(x)=a_0 x^2+a_1 x+a_2 \\
& f^{\prime}(0)=1 \\
& f^{\prime}(1)=0
\end{aligned}
\)
\(a_0, a_1, a_2\) are in A. G. P
Common difference of \(A P=1\)
Common ratio of \(G P=2\)
A.P terms \(=a, a+1, a+2\)
G.P terms \(=y, r y, r^2 y\)
\(\therefore\) AGP terms \(=a y,(a+1) r y,(a+2) r^2 y\)
\(
\begin{aligned}
& \therefore a_0=a y \\
& a_1=(a+1) r y=(a+1) 2 y \\
& a_2=(a+2) r^2 y=(a+2) 4 y
\end{aligned}
\)
Now, \(f^{\prime}(x)=2 x a_0+a_1\)
\(
\therefore f^{\prime}(0)=a_1=1
\)
\(
\begin{aligned}
& \text { and } f^{\prime}(1)=2 a_0+a_1=0 \\
& \Rightarrow 2 a_0+1=0 \\
& \Rightarrow a_0=-\frac{1}{2} \\
& \therefore a y=-\frac{1}{2} \\
& \text { and }(a+1) 2 y=1 \\
& \Rightarrow 2 a y+2 y=1 \\
& \Rightarrow 2 \times\left(-\frac{1}{2}\right)+2 y=1 \\
& \Rightarrow 2 y=+2 \\
& \Rightarrow y=+1 \\
& \therefore a=-\frac{1}{2} \\
& \therefore a_2=(a+2) 4 y \\
& =\left(-\frac{1}{2}+2\right) \times 4.1 \\
& =6 \\
& \therefore f(x)=-\frac{1}{2} x^2+x+6 \\
& \therefore f(4)=-\frac{1}{2}(4)^2+4+6 \\
& =-8+10 \\
& =2
\end{aligned}
\)

Hint

1st AP:
\(3,6,9,12 \ldots\), upto 78 terms
\(
\begin{aligned}
& t _{78}=3+(78-1) 3 \\
& =3+77 \times 3 \\
& =234
\end{aligned}
\)
2nd AP :
\(5,9,13,17, \ldots .\). upto 59 terms
\(
\begin{aligned}
& t _{59}=5+(59-1) 4 \\
& =5+58 \times 4 \\
& =237
\end{aligned}
\)
Common term’s AP :
First term \(=9\)
Common difference of first \(AP =3\)
And common difference of second AP \(=4\)
\(\therefore\) Common difference of common terms
\(
A P=\operatorname{LCM}(3,4)=12
\)
\(\therefore\) New AP \(=9,21,33, \ldots \ldots\).
\(
\begin{aligned}
& t _{ n }=9+( n -1) 12 \leq 234 \\
& \Rightarrow n \leq \frac{237}{12} \\
& \Rightarrow n=19 \\
& \therefore S_{19}=\frac{19}{2}[2.9+(19-1) 12] \\
& =19(9+108) \\
& =2223
\end{aligned}
\)

Hint

\(
S_n=\frac{n^2}{1-\frac{1}{(n+1)^2}}=\frac{n(n+1)^2}{n+2}=\left(n^2+1\right)-\frac{2}{n+2}
\)
Now \(\frac{1}{26}+\sum_{n=1}^{50}\left(S_n+\frac{2}{n+1}-n-1\right)\)
\(
\begin{aligned}
& =\frac{1}{26}+\sum_{n=1}^{50}\left\{\left(n^2-n\right)+2\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\right\} \\
& =\frac{1}{26}+\frac{50 \times 51 \times 101}{6}-\frac{50 \times 51}{2}+2\left(\frac{1}{2}-\frac{1}{52}\right) \\
& =1+25 \times 17(101-3) \\
& =41651
\end{aligned}
\)

Hint

\(
\text { Let G.P. be } a_1=a, a_2=a r, a_3=a r^2 \text {, } \lots
\)
\(
\begin{aligned}
& \because 3 a_2+a_3=2 a_4 \\
& \Rightarrow 3 a r+a r^2=2 a r^3 \\
& \Rightarrow 2 a r^2-r-3=0 \\
& \therefore r=-1 \text { or } \frac{3}{2}
\end{aligned}
\)
\(\because a_1= a >0\) then \(r \neq-1\)
Now, \(a _2+ a _4=2 a _3+1\)
\(
\begin{aligned}
& a r+a r^3=2 a r^2+1 \\
& a\left(\frac{3}{2}+\frac{27}{8}-\frac{9}{2}\right)=1 \\
& \therefore a=\frac{8}{3}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore a_2+a_4+2 a_5=a\left(r+r^3+2 r^4\right) \\
& =\frac{8}{3}\left(\frac{3}{2}+\frac{27}{8}+\frac{81}{8}\right)=40
\end{aligned}
\)

Hint

If we write the elements of \(A+A\), we can certainly find 39 distinct elements as
\(
\begin{aligned}
& 1+1,1+a_1, 1+a_2, \ldots 1 \\
& +a_{18}, 1+77, a_1+77, a_2+77, \ldots \ldots a_{18}+77,77+77
\end{aligned}
\)

It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P.

Let the common difference be ‘ \(d\) ‘.
\(
77=1+19 d \Rightarrow d=4
\)
So, \(\sum_{i=1}^{18} a_1=\frac{18}{2}\left[2 a_1+17 d\right]=9[10+68]=702\)

Hint

\(
\begin{aligned}
& T_r=\frac{r}{\left(2 r^2\right)^2+1} \\
& =\frac{r}{\left(2 r^2+1\right)^2-(2 r)^2} \\
& =\frac{1}{4} \frac{4 r}{\left(2 r^2+2 r+1\right)\left(2 r^2-2 r+1\right)} \\
& S_{10}=\frac{1}{4} \sum_{r=1}^{10}\left(\frac{1}{\left(2 r^2-2 r+1\right)}-\frac{1}{\left(2 r^2+2 r+1\right)}\right) \\
& =\frac{1}{4}\left[1-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\ldots+\frac{1}{181}-\frac{1}{221}\right] \\
& \Rightarrow S_{10}=\frac{1}{4} \cdot \frac{220}{221}=\frac{55}{221}=\frac{m}{n} \\
& \therefore m+n=276
\end{aligned}
\)

Hint

\(
\alpha_n=19^n-12^n
\)
Let equation of roots \(12 \& 19\) i.e.
\(
\begin{aligned}
& x^2-31 x+228=0 \\
& \Rightarrow(31-x)=\frac{228}{x} \text { (where } x \text { can be } 19 \text { or } 12 \text { ) }
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \frac{31 \alpha_9-\alpha_{10}}{57 \alpha_8}=\frac{31\left(19^9-12^9\right)-\left(19^{10}-12^{10}\right)}{57\left(19^8-12^8\right)} \\
& =\frac{19^9(31-19)-12^9(31-12)}{57\left(19^8-12^8\right)} \\
& =\frac{228\left(19^8-12^8\right)}{57\left(19^8-12^8\right)}=4 .
\end{aligned}
\)

Hint

\(
\begin{aligned}
& S=\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\ldots . \\
& =\sum_{r=1}^{100}\left(\frac{3^r-2^r}{3^r}\right) \\
& =100-\frac{2}{3} \frac{\left(1-\left(\frac{2}{3}\right)^{100}\right)}{1 / 3} \\
& =98+2\left(\frac{2}{3}\right)^{100} \\
& \therefore[S]=98
\end{aligned}
\)

Hint

\(A=4\)-digit numbers divisible by 3
\(
A=1002,1005, \ldots . ., 9999
\)
\(
\begin{aligned}
& 9999=1002+(n-1) 3 \\
& \Rightarrow(n-1) 3=8997 \Rightarrow n=3000
\end{aligned}
\)
\(B=4\)-digit numbers divisible by 7
\(B=1001,1008, \ldots \ldots ., 9996\)
\(
\begin{aligned}
& \Rightarrow 9996=1001+( n -1) 7 \\
& \Rightarrow n =1286
\end{aligned}
\)
\(A \cap B=1008,1029, \ldots . ., 9996\)
\(
\begin{aligned}
& 9996=1008+(n-1) 21 \\
& \Rightarrow n=429
\end{aligned}
\)
So, no divisible by either 3 or 7
\(
=3000+1286-429=3857
\)
total 4-digits numbers \(=9000\)
required numbers \(=9000-3857=5143\)

Hint

\(
\begin{aligned}
& S=\frac{7}{5}+\frac{9}{5^2}+\frac{13}{5^3}+\frac{19}{5^4}+\ldots \\
& \frac{1}{5} S=\frac{7}{5}+\frac{9}{5^3}+\frac{13}{5^4}+\ldots
\end{aligned}
\)
On subtracting
\(
\begin{aligned}
& \frac{4}{5} S=\frac{7}{5}+\frac{2}{5^2}+\frac{4}{5^3}+\frac{6}{5^4}+\ldots \\
& S=\frac{7}{14}+\frac{1}{10}\left(1+\frac{2}{5}+\frac{3}{5^2}+\ldots\right) \\
& S=\frac{7}{4}+\frac{1}{10}\left(1-\frac{1}{5}\right)^{-2} \\
& =\frac{7}{4}+\frac{1}{10} \times \frac{25}{16}=\frac{61}{32} \\
& \Rightarrow 160 S=5 \times 61=305
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 209,220,231, \ldots \ldots \ldots, 495 \\
& \text { Sum }=\frac{27}{2}(209+495)=9504
\end{aligned}
\)
Number containing 1 at unit place
\(
\begin{array}{lll}
\underline{2} & \underline{3} & \underline{1} \\
\underline{3} & \underline{4} & \underline{1} \\
\underline{4} & \underline{5} & \underline{1}
\end{array}
\)
Number containing 1 at \(10^{\text {th }}\) place
\(
\begin{array}{lll}
\underline{3} & \underline{1} & \underline{9} \\
\underline{4} & \underline{1} & \underline{8}
\end{array}
\)
Required \(=9504-(231+341+451+319+418)\)
\(
=7744
\)

Hint

\(a_1, a_2, a_3, \ldots, a_{10}\) are in AP common difference \(=-3\)
\(b_1, b_2, b_3, \ldots, b_{10}\) are in GP common ratio \(=2\)
Since, \(c_k=a_k+b_k, k=1,2,3 \ldots \ldots, 10\)
\(
\begin{aligned}
& \therefore c_2=a_2+b_2=12 \\
& c_3=a_3+b_3=13
\end{aligned}
\)
Now, \(C _3- C _2=1\)
\(
\begin{aligned}
& \Rightarrow \quad\left(a_3-a_2\right)+\left(b_3-b_2\right) \neq 1 \Rightarrow-3+\left(2 b_2-b_2\right) \neq 1 \\
& \Rightarrow \quad b_2=4 \\
& \therefore \quad a_2=8
\end{aligned}
\)
So, \(AP\) is \(11,8,5, \ldots\).
\(
\begin{aligned}
& \text { Now, } \sum_{k=1}^{10} C_k=\sum_{k=1}^{10} a_k+\sum_{k=1}^{10} b_k \\
& =\left(\frac{10}{2}\right)[22+9(-3)]+2\left(\frac{2^{10}-1}{2-1}\right) \\
& =5(22-27)+2(1023)=2046-25 \\
& =2021
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\begin{aligned}
& 2 \log _3\left(2^x-5\right)=\log _2+\log _3\left(2^x-\frac{7}{2}\right) \\
& \text { Let } 2^x=t \\
& \log _3(t-5)^2=\log _3 2\left(t-\frac{7}{2}\right) \\
& (t-5)^2=2 t-7 \\
& t^2-12 t+32=0 \\
& (t-4)(t-8)=0 \\
& \Rightarrow 2^x=4 \text { or } 2^x=8 \\
& x=2 \text { (Rejected) }
\end{aligned}\\
&\text { Or } x=3
\end{aligned}
\)

Hint

\(
l=(\underbrace{1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}}_S+\ldots)^{\log _{0.25}\left(\frac{1}{3}+\frac{1}{3^2}+\ldots\right)}
\)
\(
\begin{aligned}
& S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\ldots \\
& \frac{S}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\ldots \ldots \frac{2 x}{3}=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\ldots \\
& \frac{2 S}{3}=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\ldots \ldots \\
& S=\frac{3}{2}\left(\frac{4 / 3}{1-1 / 3}\right)=3
\end{aligned}
\)
Now, \(l=(3)^{\log _{0.25}\left(\frac{1 / 3}{1-1 / 3}\right)}\)
\(
l=3^{\log _{((1 / 4))}\left(\frac{1}{2}\right)}=3^{1 / 2}=\sqrt{3}
\)
\(
\Rightarrow l^2=3
\)

Hint

\(
2040=2^3 \times 3 \times 5 \times 17
\)
\(n\) should not be multiple of \(2,3,5\) and 17 .
\(
\begin{aligned}
& \text { Sum of all } n=(1+3+5+\ldots . .+99)-(3+9+15+21+\ldots . .+99)-(5+25+35 \\
& +55+65+85+95)-(17) \\
& =2500-\frac{17}{2}(3+99)-365-17 \\
& 2500-867-365-17 \\
& =1251
\end{aligned}
\)

Hint

\(
\frac{1}{\alpha(\alpha+1)(\alpha+2) \ldots \ldots \ldots .(\alpha+20)}=\sum_{K=0}^{20} \frac{A_k}{\alpha+k}
\)
\(
A_{14}=\frac{1}{(-14)(-13) \ldots \ldots .(-1)(1) \ldots \ldots . .(6)}=\frac{1}{14!6!}
\)
\(
A_{15}=\frac{1}{(-15)(-14) \ldots \ldots .(-1)(1) \ldots \ldots . .(5)}=\frac{1}{15!.5!}
\)
\(
A_{13}=\frac{1}{(-13) \ldots \ldots .(-1)(1) \ldots \ldots . .(7)}=\frac{1}{13!.7!}
\)
\(
\begin{aligned}
& \frac{A_{14}}{A_{13}}=\frac{1}{14!.6!} \times-13!\times 7!=\frac{-7}{14}=-\frac{1}{2} \\
& \frac{A_{15}}{A_{13}}=\frac{1}{15!\times 5!} \times-13!\times 7!=\frac{42}{15 \times 14}=\frac{1}{5} \\
& 100\left(\frac{A_{14}}{A_{13}}+\frac{A_{15}}{A_{13}}\right)^2=100\left(-\frac{1}{2}+\frac{1}{5}\right)^2=9
\end{aligned}
\)

Hint

\(
a_{n+2}=2 a_{n+1}+a_n \text {, let } \sum_{n=1}^{\infty} \frac{a_n}{8^n}=P
\)
Divide by \(8^n\) we get
\(
\begin{aligned}
& \frac{a_{n+2}}{8^n}=\frac{2 a_{n+1}}{8^n}+\frac{a_n}{8^n} \\
& \Rightarrow 64 \frac{a_{n+2}}{8^{n+2}}=\frac{16 a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}
\end{aligned}
\)
\(
\begin{aligned}
& 64 \sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}}=16 \sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}}+\sum_{n=1}^{\infty} \frac{a_n}{8^n} \\
& 64\left(P-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(P-\frac{a_1}{8}\right)+P \\
& \Rightarrow 64\left(P-\frac{1}{8}-\frac{1}{64}\right)=16\left(P-\frac{1}{8}\right)+P \\
& 64 P-8-1=16 P-2+P \\
& 47 P=7
\end{aligned}
\)

Hint

\(
\begin{aligned}
& S_n(x)=\log _a x^2+\log _a x^3+\log _a x^6+\log _a x^{11} \\
& S_n(x)=2 \log _a x+3 \log _a x+6 \log _a x+11 \log _a x+\ldots \ldots \\
& S_n(x)=\log _a x(2+3+6+11+\ldots \ldots) \\
& S_r=2+3+6+11
\end{aligned}
\)
\(
\begin{aligned}
& S_r=2+3+6+11 \\
& \therefore T_n=2+(1+3+5+\ldots \ldots+(n-1))
\end{aligned}
\)
\(
\begin{aligned}
& =2+\frac{n-1}{2}[2.1+(n-2) 2] \\
& =2+(n-1)[1+(n-2)] \\
& =n^2-2 n+3
\end{aligned}
\)
General term \(T_r=r^2-2 r+3\)
\(
\begin{aligned}
& S_n(x)=\sum_{r=1}^n \log _a x\left(r^2-2 r+3\right) \\
& S_{24}(x)=\sum_{r=1}^{24} \log _a x\left(r^2-2 r+3\right) \\
& S_{24}(x)=\log _a x \sum_{r=1}^{24}\left(r^2-2 r+3\right) \\
& 1093=4372 \log _a x
\end{aligned}
\)
\(
\begin{aligned}
& \log _a x=\frac{1}{4} \\
& x=a^{1 / 4} \ldots . .( i ) \\
& S_{12}(2 x)=\log _a(2 x) \sum_{r=1}^{12}\left(r^2-2 r+3\right) \\
& 265=530 \log _a(2 x) \\
& \log _a(2 x)=\frac{1}{2} \\
& 2 x=a^{1 / 2} \ldots \text { (ii) }
\end{aligned}
\)
From (i) and (ii), we get
\(
\begin{aligned}
& 2 a^{\frac{1}{4}}=a^{\frac{1}{2}} \\
& \Rightarrow\left(2 a^{\frac{1}{4}}\right)^4=\left(a^{\frac{1}{2}}\right)^4 \\
& \Rightarrow 16 a=a^2 \\
& \Rightarrow a=16
\end{aligned}
\)

Hint

\(a^2=\frac{b}{16}\) and \(\frac{2}{b}=\frac{1}{a}+6\)
Solving, we get \(a=\frac{1}{12}\) or \(a=-\frac{1}{4}\) [rejected]
if \(a=\frac{1}{12} \Rightarrow b=\frac{1}{9}\)
\(
\therefore 72(a+b)=72\left(\frac{1}{12}+\frac{1}{9}\right)=14
\)

Hint

A.P. from the set will be \(11,16,21,26 \ldots .\).
G.P. from the set will be \(4,8,16,32,64,128,256,512,1024,2048,4096,8192 \ldots .\).

So common terms are \(16,256,4096\).

Hint

Let \(N\) be the four digit number
\(
\operatorname{gcd}( N , 18)=3
\)

Hence \(N\) is an odd integer which is divisible by 3 but not by 9 .
4 digit odd multiples of 3
1005,1011, \(\ldots\) \(9999 \rightarrow 1500\)
4 digit odd multiples of 9
1017,1035, \(\ldots\) \(9999 \rightarrow 500\)

Hence number of such \(N =1000\)

Hint

Given, \(4 x^2-9 x+5=0\)
\(
\begin{aligned}
& \Rightarrow(x-1)(4 x-5)=0 \\
& \Rightarrow \text { A. M. }=\frac{5}{4}, \text { G. M. }=1 \text { (As A. M. } \geq \text { G. M) }
\end{aligned}
\)
Again, for the series
\(
-16,8,-4,2 \ldots
\)
\(
\begin{aligned}
& p^{t h} \text { term } t_p=-16\left(\frac{-1}{2}\right)^{p-1} \\
& q^{\text {th }} \text { term } t_p=16\left(\frac{-1}{2}\right)^{q-1} \\
& \text { Now, A. M. }=\frac{t_p+t_q}{2}=\frac{5}{4} \& \text { G. M. }=\sqrt{t_p t_q}=1 \\
& \Rightarrow 16^2\left(-\frac{1}{2}\right)^{p+q-2}=1 \\
& \Rightarrow(-2)^8=(-2)^{(p+q-2)} \\
& \Rightarrow p+q=10
\end{aligned}
\)

Hint

Let the terms are \(a, a r, a r^2, a r^3\)
\(
\begin{aligned}
& a+a r+a r^2+a r^3=\frac{65}{12} \dots(1)\\
& \frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}+\frac{1}{a r^3}=\frac{65}{18} \\
& \frac{1}{a}\left(\frac{r^3+r^2+r+1}{r^3}\right)=\frac{65}{18} \ldots \ldots \ldots(2)
\end{aligned}
\)
Doing \(\frac{(1)}{(2)}\),
\(
a^2 r^3=\frac{18}{12}=\frac{3}{2}
\)
Also given, \(a^3 r^3=1 \Rightarrow a\left(\frac{3}{2}\right)=1 \Rightarrow a=\frac{2}{3}\)
\(
\begin{aligned}
& \frac{4}{9} r^3=\frac{3}{2} \Rightarrow r^3=\frac{3^3}{2^3} \Rightarrow r=\frac{3}{2} \\
& \alpha=a r^2=\frac{2}{3} \cdot\left(\frac{3}{2}\right)^2=\frac{3}{2} \\
& 2 \alpha=3
\end{aligned}
\)

Hint

Given \(m\) arithmetic means (A.Ms) present between 3 and 243
\(\therefore\) Common difference, \(d=\frac{b-a}{m+1}=\frac{240}{m+1}\)
\(\therefore\) 4th A.M. \(= a +4 d\)
\(
=3+4 \times \frac{240}{m+1}
\)
Also there are 3 G.M between 3 and 243
\(\therefore\) Common ratio \(( r )=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}\)
where \(n =\) number of G.M inserted.
\(
\therefore r=\left(\frac{243}{3}\right)^{\frac{1}{3+1}}=3
\)
Given,
\(
\begin{aligned}
& 4^{\text {th }} A . M=2^{\text {nd }} G . M \\
& \Rightarrow 3+4 \times \frac{240}{m+1}=3(3)^2 \\
& \Rightarrow \frac{960}{m+1}=24 \\
& \Rightarrow m=39
\end{aligned}
\)

Hint

Given, \((0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\ldots . \text { to } \infty\right)}\)
As sum of GP upto infinity \(=\frac{a}{1-r}\)
\(
\therefore \frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \infty \infty=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}
\)
\(
\begin{aligned}
& \therefore(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\ldots . \text { to } \infty\right)} \\
& =(0.16)^{\log _{2.5}\left(\frac{1}{2}\right)} \\
& =\left(\frac{16}{100}\right)^{\log _{2.5}\left(\frac{1}{2}\right)} \\
& =\left(\frac{4}{10}\right)^{\log _{2.5}\left(\frac{1}{2}\right)} \\
& =\left[\left(\frac{10}{4}\right)^{-2}\right]^{\log _{2.5}\left(\frac{1}{2}\right)} \\
& =\left[(2.5)^{-2}\right]^{\log _{2.5}\left(\frac{1}{2}\right)} \\
& =(2.5)^{-2 \log _{2.5}\left(\frac{1}{2}\right)} \\
& =\left(\frac{1}{2}\right)^{-2}=4 \\
&
\end{aligned}
\)

Hint

First A.P. is \(3,7,11,15,19,23, \ldots . .407\)
\(
d _1=4
\)
Second A.P. is \(2,9,16,23, \ldots . .709\)
\(
d_2=7
\)
First common term \(=23\)

Common difference of new A.P using the common terms of the two given A.P’s is \(d =\) L.C.M. \((4,7)=28\)
Last term \(\leq 407\)
\(
\begin{aligned}
& \Rightarrow 23+(n-1)(28) \leq 407 \\
& \Rightarrow n \leq 14.7 \\
& \therefore n=14
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sum_{n=1}^7 \frac{n(n+1)(2 n+1)}{4} \\
& =\frac{1}{4} \sum_{n=1}^7\left(2 n^3+3 n^2+n\right) \\
& =\frac{1}{2} \sum_{n=1}^7 n^3+\frac{3}{4} \sum_{n=1}^7 n^2+\frac{1}{4} \sum_{n=1}^7 n \\
& =\frac{1}{2}\left(\frac{7(7+1)}{2}\right)^2+\frac{3}{4}\left(\frac{7(7+1)(14+1)}{6}\right)+\frac{1}{4} \frac{7(8)}{2} \\
& =(49)(8)+(15 \times 7)+(7) \\
& =392+105+7=504
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sum_{k=1}^{20}(1+2+3+\ldots+k) \\
& =\sum_{k=1}^{20} \frac{k(k+1)}{2} \\
& =\sum_{k=1}^{20} \frac{k^2}{2}+\sum_{k=1}^{20} \frac{k}{2} \\
& =\frac{1}{2} \times \frac{20 \times 21 \times 41}{6}+\frac{1}{2} \times \frac{20 \times 21}{2} \\
& =1540
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \sum_{n=0}^{\infty} a r^n=57 \quad \Rightarrow \frac{a}{1-r}=57 \dots(i) \\
& \sum_{n=0}^{\infty} a^3 r^{3 n}=9747 \quad \Rightarrow \frac{a^3}{1-r^3}=9747 \dots(ii) \\
& \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\
& \Rightarrow \quad \frac{(1-r)\left(1+r+r^2\right)}{(1-r)^3}=19 \\
& \Rightarrow \quad 18 r^2-39 r+18=0 \\
& \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\
& \therefore \quad a=19 \\
& \quad a+18 r \\
& \quad=19+12=31
\end{aligned}
\)

Hint

\(
\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5
\)
Multiply and divide by \(d\)
\(
\begin{aligned}
& \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\
& \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots+\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\
& \frac{1}{d}\left[1-\frac{1}{1+10 d}\right]=5 \\
& \frac{1}{d}\left[\frac{1+10 d-1}{1+10 d}\right]=5 \\
& \frac{10}{1+10 d}=5 \\
& 1+10 d=2 \\
& d=\frac{1}{10} \\
& 50 d=50 \times \frac{1}{10}=5
\end{aligned}
\)

Hint

Let’s denote the first term of the geometric progression by \(a\) and the common ratio by \(r\). The terms of the geometric progression can be written as follows:
First term: \(a\)
Second term: \(a r\)
Third term: \(\left(a r^2\right)\)
Fourth term: \(\left(a r^3\right)\)
Fifth term: \(\left(a r^4\right)\)
Sixth term: \(\left(a r^5\right)\)
Eighth term: \(\left(a r^7\right)\)
We are given two key pieces of information:
1. The sum of the second and sixth terms is \(\frac{70}{3}\) :
\(
a r+a r^5=\frac{70}{3}
\)
2. The product of the third and fifth terms is 49 :
\(
\begin{aligned}
& \left(a r^2\right) \cdot\left(a r^4\right)=49 \\
& a^2 r^6=49 \\
& a^2=\frac{49}{r^6} \\
& a=\frac{7}{r^3}
\end{aligned}
\)
Substituting \(a=\frac{7}{r^3}\) into the first equation:
\(
\begin{aligned}
& \frac{7}{r^3} \cdot r+\frac{7}{r^3} \cdot r^5=\frac{70}{3} \\
& \frac{7 r}{r^3}+\frac{7 r^5}{r^3}=\frac{70}{3} \\
& \frac{7}{r^2}+\frac{7 r^2}{1}=\frac{70}{3}
\end{aligned}
\)
Let \(x=r^2\). Then:
\(
\frac{7}{x}+7 x=\frac{70}{3}
\)
Multiply through by \(3 x\) to clear the denominator:
\(
21+21 x^2=70 x
\)
Rearrange into a standard quadratic equation:
\(
21 x^2-70 x+21=0
\)
Divide by 7 to simplify:
\(
3 x^2-10 x+3=0
\)
Solve this quadratic equation using the quadratic formula:
\(
x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}
\)
Where \(a=3, b=-10\), and \(c=3\). Thus:
\(
\begin{aligned}
& x=\frac{10 \pm \sqrt{100-36}}{6} \\
& x=\frac{10 \pm \sqrt{64}}{6} \\
& x=\frac{10 \pm 8}{6} \\
& x=3 \text { or } x=\frac{1}{3}
\end{aligned}
\)
Since \(x=r^2\) and \(r\) is positive, we get \(r=\sqrt{3}\) or \(r=\frac{1}{\sqrt{3}}\). We need to choose the value that results in positive, increasing terms:
If \(r=\sqrt{3}\) :
\(
a=\frac{7}{r^3}=\frac{7}{(\sqrt{3})^3}=\frac{7}{3 \sqrt{3}}=\frac{7}{3} \cdot \frac{1}{\sqrt{3}}=\frac{7 \sqrt{3}}{9}
\)
Now we can determine the sum of the 4 th, 6 th, and 8th terms:
The 4th term is: \(a r^3=\frac{7 \sqrt{3}}{9} \cdot 3 \sqrt{3}=7\)
The 6 th term is: \(a r^5=\frac{7 \sqrt{3}}{9} \cdot 9 \sqrt{3}=21\)
The 8th term is: \(a r^7=\frac{7 \sqrt{3}}{9} \cdot 27(\sqrt{3})=49\)
Adding these together:
\(
(4 t h+6 \text { th }+8 \text { thterms })=7+21+63=91
\)
Therefore, the sum of the 4 th, 6th, and 8th terms is 91 .

Hint

\(\triangle A B C\) is an equilateral triangle having side \(=a\) unit
Now, perimeter \(=\) sum of all sides \(=3 a\)
\(
\text { Area }=\frac{\sqrt{3}}{4} a^2
\)
Now, \(\ln \triangle D E F, D E=\frac{a}{2}=E F=D F\)
Perimeter \(=3 \times \frac{a}{2}=\frac{3 a}{2}\)
Area \(=\frac{\sqrt{3}}{4} \times\left(\frac{a}{2}\right)^2=\frac{\sqrt{3} a^2}{16}\)
Now, \(P=3 a+\frac{3 a}{2}+\frac{3 a}{4}+\cdots\)
\(
\begin{aligned}
& Q=\frac{\sqrt{3}}{4} a^2+\frac{\sqrt{3}}{16} a^2+\frac{\sqrt{3}}{64} a^2+\cdots \\
& P=\frac{3 a}{1-\frac{1}{2}}=3 a \times 2 \Rightarrow P=6 a \dots(i) \\
& Q=\frac{\frac{\sqrt{3}}{4} a^2}{1-\frac{1}{4}}=\frac{4}{3} \times \frac{\sqrt{3}}{4} a^2 \quad Q=\frac{\sqrt{3}}{3} a^2 \dots(ii)
\end{aligned}
\)
From equation (i) & (ii)
\(
\begin{aligned}
& P=6 a \\
& Q=\frac{\sqrt{3}}{3} a^2 \\
& Q=\frac{\sqrt{3}}{3} \times \frac{P^2}{36} \\
& P^2=36 \sqrt{3} Q
\end{aligned}
\)

Hint

To determine the value of \(m\), we need to formulate the problem using some basic concepts of arithmetic progression and work. Let’s first understand the nature of the problem:
Initially, there are \(m\) computers, and it is estimated that with these \(m\) computers, the assignment can be completed in 17 days.
However, due to the crash of 4 computers every day starting from the second day onward, the total time taken extends by 8 days, making it 25 days in total.
To begin with, let’s define the total work ( \(W\) ) in terms of the number of computers and days:
The total work \(( W )\) is given by:
The amount of work completed each day with \(m\) computers for 17 days:
\(
W=17 m
\)
When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are \(m\) computers. On the second day, there are \(m -4\) computers, on the third day, there are \(m -8\) computers, and so on. We need to sum this series until 25 days are completed.
This can be formulated as:
Total work done over 25 days with decrement in the number of computers:
\(
W=m+(m-4)+(m-8)+\ldots+[m-4 \times(n-1)]
\)
where \(n\) is the number of days. Here, \(n=25\).
Notice that we form an arithmetic series where the first term (a) is \(m\) and the common difference (d) is -4 . The sum of the first \(n\) terms of an arithmetic series is:
\(
S_n=\frac{n}{2}[2 a+(n-1) d]
\)
Plugging in the values:
\(
\begin{aligned}
& S_{25}=\frac{25}{2}[2 m+(25-1)(-4)] \\
& S_{25}=\frac{25}{2}[2 m-96] \\
& S_{25}=\frac{25}{2}[2 m-96]=25(m-48)
\end{aligned}
\)
This work should be equivalent to the work calculated earlier, so:
\(
17 m=25(m-48)
\)
Solving for \(m\) :
\(
\begin{aligned}
& 17 m=25 m-1200 \\
& 8 m=1200 \\
& m=150
\end{aligned}
\)

Hint

To determine the least value of \(K[latex] for which the terms [latex]4^{1+x}+4^{1-x}, \frac{ K }{2}, 16^x+16^{-x}\) form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.
For three numbers to be in an arithmetic progression, the middle term must be the average of the other two terms. Therefore, we can write:
\(
\frac{4^{1+x}+4^{1-x}+16^x+16^{-x}}{2}=\frac{K}{2}
\)
First, simplify each term individually:
1. Consider \(4^{1+x}+4^{1-x}\) :
\(
4^{1+x}=4 \cdot 4^x=4 \cdot\left(2^2\right)^x=4 \cdot 2^{2 x}=4 \cdot 2^{2 x}
\)
and \(4^{1-x}=4 \cdot 4^{-x}=4 \cdot\left(2^2\right)^{-x}=4 \cdot 2^{-2 x}=4 \cdot 2^{-2 x}\)
Thus,
\(
4^{1+x}+4^{1-x}=4 \cdot 2^{2 x}+4 \cdot 2^{-2 x}=4\left(2^{2 x}+2^{-2 x}\right)
\)
2. Consider \(16^x+16^{-x}\) :
\(
16^x=\left(2^4\right)^x=2^{4 x}
\)
and \(16^{-x}=\left(2^4\right)^{-x}=2^{-4 x}\)
Thus,
\(
16^x+16^{-x}=2^{4 x}+2^{-4 x}
\)
3. Combine the terms and set up the equation:
\(
\frac{4\left(2^{2 x}+2^{-2 x}\right)+2^{4 x}+2^{-4 x}}{2}=\frac{K}{2}
\)
Multiply both sides by 2 :
\(
4\left(2^{2 x}+2^{-2 x}\right)+2^{4 x}+2^{-4 x}=K
\)
To find the least value of \(K\), let’s assume \(x=0\) (since \(x\) can range over non-negative values):
For \(x=0\) :
\(
4\left(2^{2 \cdot 0}+2^{-2 \cdot 0}\right)+2^{4 \cdot 0}+2^{-4 \cdot 0}
\)
This simplifies to:
\(
\begin{aligned}
& 4\left(2^0+2^0\right)+2^0+2^0 \\
& =4(1+1)+1+1 \\
& =4 \cdot 2+1+1 \\
& =8+1+1 \\
& =10
\end{aligned}
\)
Therefore, the least value of \(K\) that ensures the values form an arithmetic progression is 10 .

Hint

\(
\begin{aligned}
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m \\
& \text { and } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n \\
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \\
& =\frac{1}{\sqrt{1}+\sqrt{2}} \times \frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\
& \quad+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \times \frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}} \\
& =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{100}-\sqrt{99} \\
& =\sqrt{100}-\sqrt{1} \\
& =10-1 \\
& \Rightarrow m=9
\end{aligned}
\)
and \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n\)
\(
\begin{aligned}
& \frac{2-1}{1 \times 2}+\frac{3-2}{2 \times 3}+\ldots+\frac{100-99}{100 \times 99}=n \\
& \Rightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots+\frac{1}{99}-\frac{1}{100}=n \\
& \Rightarrow n=1-\frac{1}{100} \\
& \Rightarrow n=\frac{99}{100} \\
& (m, n)=\left(9, \frac{99}{100}\right)
\end{aligned}
\)
Satisfies the line \(11 x-100 y=0\)

Hint

\(
\begin{aligned}
& \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101} \\
& \Rightarrow \frac{\sum_{n=1}^{100} n(n+1)^2}{\sum_{n=1}^{100} n^2(n+1)} \\
& \Rightarrow \frac{\sum_{n=1}^{100} n^3+2 n^2+n}{\sum_{n=1}^{100} n^3+n^2}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{\left(\frac{100(101)}{2}\right)^2+\frac{2 \cdot 100(101)(201)}{6}+\frac{100(101)}{2}}{\left(\frac{100(101)}{2}\right)^2+\frac{100(101)(201)}{6}} \\
& =\frac{300(101)+4(201)+6}{300(101)+2(201)}=\frac{5185}{5117}=\frac{305}{301}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 2 b=a+c \dots(1) \\
& b^2=(a+1)(c+3) \dots(2) \\
& \frac{a+b+c}{3}=8 \dots(3) \\
& \Rightarrow \frac{3 b}{3}=8 \\
& \quad b=8 \\
& \Rightarrow \quad a c+3 a+c+3=64 \\
& 3 a+c+a c=61 \dots(4) \\
& a+c=16 \\
& c=16-a
\end{aligned}
\)
from equation (4)
\(
\begin{aligned}
& 3 a+16-a+a(16-a)=61 \\
& \Rightarrow \quad(a-15)(a-3)=0 \\
& \quad a=15(a>10) \\
& \Rightarrow \quad a=15, b=8, c=1 \\
& \left((a \cdot b \cdot c)^{\frac{1}{3}}\right)^3=15 \times 8 \times 1=120
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Let } p=2 r, q=2 r^2 \\
& T_7=2, T_8=2 r, T_{13}=2 r^2 \\
& d=2 r-2=2(r-1) \\
& 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \\
& \Rightarrow r^2-6 r+5=0 \\
& \Rightarrow(r-1)(r-5)=0 \\
& \therefore r=1,5 \\
& r=1 \text { (rejected) as } q \neq 2 \\
& \therefore r=5 \\
& 5^{\text {th }} \text { term of G.P }=2 . r^4=2.5^4 \\
& \text { Let } 1^{\text {st }} \text { term of A.P } b a=a, d=8 \\
& 2=a+(6)(8) \Rightarrow a=-46 \\
& n ^{\text {th }} \text { term of A.P }=-46+(n-1) 8=8 n-54 \\
& 2.5^4=8 n-54 \\
& \Rightarrow 1250+54=8 n \\
& \Rightarrow n=\frac{1304}{8}=163
\end{aligned}
\)

Hint

To solve this problem, we will start by using the properties of an arithmetic progression (AP).
The sum of the first \(n\) terms of an AP can be calculated using the formula: \(S_n=\frac{n}{2}(2 a+(n-1) d)\) where \(S_n\) is the sum of the first \(n\) terms, \(a\) is the first term, and \(d\) is the common difference between the terms.
Given the information: \(S_{10}=390\)
We can plug \(n=10\) into the sum formula to get:
\(
\begin{aligned}
& S_{10}=\frac{10}{2}(2 a+(10-1) d) \\
& 390=5(2 a+9 d) \\
& 390=10 a+45 d \\
& 78=2 a+9 d \quad \ldots \ldots \ldots .(1)
\end{aligned}
\)
Next, we’re given the ratio of the tenth term \(\left(T_{10}\right)\) to the fifth term \(\left(T_5\right): \frac{T_{10}}{T_5}=\frac{15}{7}\)
The \(n\)th term of an AP is given by:
\(
T_n=a+(n-1) d
\)
So, for the tenth term: \(T_{10}=a+(10-1) d=a+9 d\)
And for the fifth term:
\(
T_5=a+(5-1) d=a+4 d
\)
Now we can write the ratio as:
\(
\begin{aligned}
& \frac{a+9 d}{a+4 d}=\frac{15}{7} \\
& 7(a+9 d)=15(a+4 d) \\
& 7 a+63 d=15 a+60 d \\
& 63 d-60 d=15 a-7 a \\
& 3 d=8 a \quad \ldots \ldots \ldots(2)
\end{aligned}
\)
Now we have two equations (1) and (2):
\(
\begin{aligned}
& 78=2 a+9 d \\
& 3 d=8 a
\end{aligned}
\)
We can solve these equations simultaneously.
From equation (2):
\(
d=\frac{8}{3} a
\)
Plugging this back into (1):
\(
\begin{aligned}
& 78=2 a+9\left(\frac{8}{3} a\right) \\
& 78=2 a+24 a \\
& 78=26 a \\
& a=3
\end{aligned}
\)
Now we can find \(d\) :
\(
\begin{aligned}
& d=\frac{8}{3} a \\
& d=\frac{8}{3} \times 3 \\
& d=8
\end{aligned}
\)
Now we can find \(S_{15}\) and \(S_5\) using the formula for the sum of an AP.
For \(S_{15}\) :
\(
\begin{aligned}
& S_{15}=\frac{15}{2}(2 \cdot 3+(15-1) \cdot 8) \\
& S_{15}=\frac{15}{2}(6+14 \cdot 8) \\
& S_{15}=\frac{15}{2}(6+112) \\
& S_{15}=\frac{15}{2} \cdot 118 \\
& S_{15}=15 \cdot 59 \\
& S_{15}=885
\end{aligned}
\)
For \(S_5\) :
\(
\begin{aligned}
& S_5=\frac{5}{2}(2 \cdot 3+(5-1) \cdot 8) \\
& S_5=\frac{5}{2}(6+4 \cdot 8) \\
& S_5=\frac{5}{2}(6+32) \\
& S_5=\frac{5}{2} \cdot 38 \\
& S_5=5 \cdot 19 \\
& S_5=95
\end{aligned}
\)
The difference \(S_{15}-S_5\) is:
\(
\begin{aligned}
& S_{15}-S_5=885-95 \\
& S_{15}-S_5=790
\end{aligned}
\)

Hint

Since \(3, a, b, c\) are in arithmetic progression (A.P.), the common difference can be calculated using the term \(a\) (the second term) as follows:
\(
d=a-3
\)
The nth term of an A.P. is given by the formula:
\(
T_n=a+(n-1) d
\)
So, using this formula, we can express \(b\) and \(c\) in terms of \(a\) and \(d\) :
\(
\begin{aligned}
& b=a+d \\
& c=a+2 d
\end{aligned}
\)
Substituting \(d=a-3\) into these expressions:
\(
\begin{aligned}
& b=a+(a-3) \\
& c=a+2(a-3)
\end{aligned}
\)
Therefore:
\(
\begin{aligned}
& b=2 a-3 \\
& c=3 a-6
\end{aligned}
\)
Now, let’s consider that \(3, a-1, b+1, c+9\) are in geometric progression (G.P.). For terms in a G.P., the ratio (common ratio, r) between consecutive terms is constant. So:
\(
\frac{a-1}{3}=\frac{b+1}{a-1}=\frac{c+9}{b+1}
\)
Now, we will establish the relation between the terms using the property of G.P.:
\(
\begin{aligned}
& \frac{a-1}{3}=\frac{b+1}{a-1} \\
& (a-1)^2=3(b+1) \\
& a^2-2 a+1=3 b+3
\end{aligned}
\)
Substituting \(b=2 a-3\), we get:
\(
\begin{aligned}
& a^2-2 a+1=3(2 a-3)+3 \\
& a^2-2 a+1=6 a-9+3 \\
& a^2-8 a+7=0
\end{aligned}
\)
Solving this quadratic equation:
\(
(a-7)(a-1)=0
\)
Hence, \(a=7\) or \(a=1\). However, if \(a=1\), the terms \(3, a-1, b+1, c+9\) cannot form a G.P. as it would involve division by zero. Therefore, \(a=7\). We use this value to find \(b\) and \(c\) :
\(
\begin{aligned}
& b=2 a-3=2(7)-3=14-3=11 \\
& c=3 a-6=3(7)-6=21-6=15
\end{aligned}
\)
Now we can find the arithmetic mean \((A)\) of \(a, b\), and \(c\) :
\(
\begin{aligned}
& A=\frac{a+b+c}{3} \\
& A=\frac{7+11+15}{3} \\
& A=\frac{33}{3} \\
& A=11
\end{aligned}
\)
Hence, the arithmetic mean of \(a, b\), and \(c\) is 11 , which corresponds to Option D.

Hint

\(
\begin{aligned}
& 1+d, \quad 1+7 d, 1+43 d \text { are in GP } \\
& (1+7 d)^2=(1+d)(1+43 d) \\
& 1+49 d^2+14 d=1+44 d+43 d^2 \\
& 6 d^2-30 d=0 \\
& d=5 \\
& S_{20}=\frac{20}{2}[2 \times 1+(20-1) \times 5] \\
& \quad=10[2+95] \\
& \quad=970
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\
& f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\
& f(1)=0 \\
& \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\
& \alpha=\frac{c+a-2 b}{a+b-2 c} \\
& \text { If, }-1<\alpha<0 \\
& -1<\frac{c+a-2 b}{a+b-2 c}<0 \\
& b+c<2 a \text { and } b>\frac{a+c}{2}
\end{aligned}
\)
therefore, \(b\) cannot be G.M. between \(a\) and \(c\).
\(
\begin{aligned}
& \text { If, } 0<\alpha<1 \\
& 0<\frac{c+a-2 b}{a+b-2 c}<1 \\
& b>c \text { and } b<\frac{a+c}{2}
\end{aligned}
\)
Therefore, \(b\) may be the G.M. between \(a\) and \(c\).

Hint

General term of the sequence,
\(
\begin{aligned}
& T _{ r }=\frac{ r }{1-3 r ^2+ r ^4} \\
& T _{ r }=\frac{ r }{ r ^4-2 r ^2+1- r ^2} \\
& T _{ r }=\frac{ r }{\left( r ^2-1\right)^2- r ^2} \\
& T _{ r }=\frac{ r }{\left( r ^2- r -1\right)\left( r ^2+ r -1\right)} \\
& T _{ r }=\frac{\frac{1}{2}\left[\left( r ^2+ r -1\right)-\left( r ^2- r -1\right)\right]}{\left( r ^2- r -1\right)\left( r ^2+ r -1\right)} \\
&=\frac{1}{2}\left[\frac{1}{ r ^2- r -1}-\frac{1}{ r ^2+ r -1}\right]
\end{aligned}
\)
Sum of 10 terms,
\(
\sum_{ r =1}^{10} T _{ r }=\frac{1}{2}\left[\frac{1}{-1}-\frac{1}{109}\right]=\frac{-55}{109}
\)

Hint

The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value. We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified to be equal.

For the first GP, with first term \(a\) and third term \(b\) :
The third term is given by \(t_3=a r^2=b\), leading to \(r^2=\frac{b}{a}\).
The eleventh term is \(t_{11}=a r^{10}=a\left(\frac{b}{a}\right)^5\).
For the second GP, with first term \(a\) and fifth term \(b\) :
The fifth term is \(T_5=a r^4=b\), yielding \(r^4=\frac{b}{a}\) and \(r=\left(\frac{b}{a}\right)^{1 / 4}\). The \(p^{\text {th }}\) term is thus \(T_p=a r^{p-1}=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}\).

Equating the eleventh term of the first GP to the \(p ^{\text {th }}\) term of the second GP gives: \(a\left(\frac{b}{a}\right)^5=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}\).
Simplifying, we find that \(5=\frac{p-1}{4}\), leading to \(p=21\), which is the solution.

Hint

\(
\begin{aligned}
& S _{20}=\frac{20}{2}[2 a +19 d ]=790 \\
& 2 a +19 d =79 \\
& S _{10}=\frac{10}{2}[2 a +9 d ]=145 \\
& 2 a +9 d =29
\end{aligned}
\)
From (1) and (2) \(a=-8, d=5\)
\(
\begin{aligned}
& S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\
& =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\
& =405-10 \\
& =395
\end{aligned}
\)

Hint

\(\log _e a, \log _e b, \log _e c\) are in A.P.
\(
\therefore b ^2= ac \ldots . . \text { (i) }
\)
Also
\(\log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right)\) are in A.P.
\(
\begin{aligned}
& \left(\frac{2 b}{3 c}\right)^2=\frac{a}{2 b} \times \frac{3 c}{a} \\
& \frac{b}{c}=\frac{3}{2}
\end{aligned}
\)
Putting in eq. (i) \(b^2=a \times \frac{2 b}{3}\)
\(
\begin{aligned}
& \frac{a}{b}=\frac{3}{2} \\
& a: b: c=9: 6: 4
\end{aligned}
\)

Hint

Let \(r^{\prime}\) th term of the GP be \(a^{n-1}\). Given,
\(
\begin{aligned}
& 2 a_r=a_{r+1}+a_{r+2} \\
& 2 a r^{n-1}=a r^n+a r^{n+1} \\
& \frac{2}{r}=1+r \\
& r^2+r-2=0
\end{aligned}
\)
Hence, we get, \(r=-2\) (as \(r \neq 1)\)
So, \(S _{20}- S _{18}=\) (Sum upto 20 terms) – (Sum upto 18 terms) \(= T _{19}+ T _{20}\)
\(
T _{19}+ T _{20}=\operatorname{ar}^{18}(1+ r )
\)
Putting the values \(a =\frac{1}{8}\) and \(r =-2\);
we get \(T_{19}+T_{20}=-2^{15}\)

Hint

\(
\begin{aligned}
& a+a r+a r^2+a r^3+\ldots+a r^{63} \\
& =7\left(a+a r^2+a r^4 \ldots+a r^{62}\right) \\
& \Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2} \\
& r=6
\end{aligned}
\)

Hint

Step 1: Find the required product of terms of A.P.
Let \(a\) be the first term and \(d\) be the common difference of the AP.
Given 6 th term \(=2\)
\(
\begin{aligned}
\Rightarrow & a+5 d=2 \\
\Rightarrow & a=2-5 d
\end{aligned}
\)
Let the product of \(a_1 a_4 a_5\) is \(P\)
\(
\begin{aligned}
P & =a(a+3 d)(a+4 d) \\
& =(2-5 d)(2-5 d+3 d)(2-5 d+4 d) \\
& =(2-5 d)(2-2 d)(2-d) \\
& =2(1-d)(2-d)(2-5 d) \\
\therefore & P=2\left(-5 d^3+17 d^2-16 d+4\right)
\end{aligned}
\)
Step 2: Find the minimum value of the above product.
For minimum value of \(P, P \prime(d)=0\)
Differentiate \(P\) w.r.t. \(d\), we get
\(
\begin{aligned}
& P \prime(d)=2\left(-15 d^2+34 d-16\right) \\
& \therefore 2\left(-15 d^2+34 d-16\right)=0 \\
& \Rightarrow \quad\left(-15 d^2+34 d-16\right)=0 \\
& \Rightarrow \quad(3 d-2)(5 d-8)=0 \\
& \Rightarrow \quad d=\frac{2}{3} \text { or } d=\frac{8}{5} \\
&
\end{aligned}
\)
Again differentiate w.r.t. \(d\), we get
\(
P \prime \prime(d)=2(-30 d+34)
\)
Now at \(d=\frac{2}{3}\)
\(
P \prime \prime(d)=28 \text { [ positive ] }
\)
and at \(d=\frac{8}{5}\)
\(
P \prime \prime(d)=-28[\text { negative }]
\)
We get minimum value when \(P \prime \prime(d)>0\).
So \(d=\frac{2}{3}\) gives the least value.

Hint

\(
20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}
\)
This is A.P. with common difference
\(
\begin{aligned}
& d_1=-1+\frac{1}{4}=-\frac{3}{4} \\
& -129 \frac{1}{4}, \ldots \ldots \ldots \ldots \ldots 19 \frac{1}{4}, 20
\end{aligned}
\)
This is also A.P. \(a =-129 \frac{1}{4}\) and \(d =\frac{3}{4}\)
Required term \(=\)
\(
\begin{aligned}
& -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\
& =-129-\frac{1}{4}+15-\frac{3}{4}=-115
\end{aligned}
\)

Hint

\(4,9,14,19, \ldots\), up to \(25^{\text {th }}\) term
\(
T _{25}=4+(25-1) 5=4+120=124
\)
\(3,6,9,12, \ldots\), up to \(37^{\text {th }}\) term
\(
T _{37}=3+(37-1) 3=3+108=111
\)
Common difference of \(I ^{\text {st }}\) series \(d _1=5\)
Common difference of \(II ^{\text {nd }}\) series \(d _2=3\)
First common term \(=9\), and their common difference \(=15\left( LCM\right.\) of \(d _1\) and \(\left.d _2\right)\) then common terms are \(9,24,39,54,69,84,99\)

Hint

Now, we have the following relations :
Arithmetic progression :
Since \(A_1\) and \(A_2\) are arithmetic means between \(a\) and \(b\), we can say that \(a, A_1, A_2\), and \(b\) are in an arithmetic progression. This means there are three equal intervals between \(a\) and \(b\), which are represented by the common difference \(d\).
To find the value of \(d\), we can use the following equation:
\(
b-a=3 d
\)
From this equation, we can find the value of \(d\) :
\(
\begin{aligned}
& d=\frac{b-a}{3} \\
& A_1=a+\frac{b-a}{3}=\frac{2 a+b}{3} \\
& A_2=\frac{a+2 b}{3} \\
& A_1+A_2=a+b
\end{aligned}
\)
Geometric progression :
\(a, G_1, G_2, G_3, b\) are in G.P.
\(r=\left(\frac{b}{a}\right)^{\frac{1}{4}}\)
\(G_1=\left(a^3 b\right)^{\frac{1}{4}}\)
\(G_2=\left(a^2 b^2\right)^{\frac{1}{4}}\)
\(G_3=\left(a b^3\right)^{\frac{1}{4}}\)
We have the expression :
\(
G_1^4+G_2^4+G_3^4+G_1^2 G_3^2=a^3 b+a^2 b^2+a b^3+\left(a^3 b\right)^{\frac{1}{2}} \cdot\left(a b^3\right)^{\frac{1}{2}}
\)
Simplify the expression :
\(
a^3 b+a^2 b^2+a b^3+a b\left(a^2 b^2\right)
\)
Factor out \(a b\) :
\(
a b\left(a^2+a b+b^2+a^2 b^2\right)
\)
Combine the terms :
\(
a b\left(a^2+2 a b+b^2\right)
\)
Rewrite the expression using the sum of squares :
\(
a b(a+b)^2
\)
Now, recall that \(A_1+A_2=a+b\). Substitute this into the expression:
\(
G_1 \cdot G_3 \cdot\left(A_1+A_2\right)^2
\)

Hint

Given the conditions :
\(
\begin{aligned}
& a_6+a_8=2 \Rightarrow a r^5+a r^7=2 \\
& a_3 \cdot a_5=\frac{1}{9} \Rightarrow a^2 \cdot r^2 \cdot r^4=\frac{1}{9} \Rightarrow a r^3=\frac{1}{3}
\end{aligned}
\)
From this, we can form the equation \(\frac{r^2}{3}+\frac{r^4}{3}=2\), which simplifies to \(r^4+r^2=6\).
This can be factored to give \(\left(r^2-2\right)\left(r^2+3\right)=0\), yielding \(r^2=2\) (since \(r^2\) cannot be -3 for real \(r)\).
So, we have \(r=\sqrt{2}\).
Substituting \(r=\sqrt{2}\) into the equation \(a r=\frac{1}{6}\), we get \(a=\frac{1}{6 \sqrt{2}}\).
Now, we find the value of \(6\left(a_2+a_4\right)\left(a_4+a_6\right)\) :
\(
\begin{aligned}
& 6\left(a_2+a_4\right)\left(a_4+a_6\right)=6\left(a r+a r^3\right)\left(a r^3+a r^5\right) \\
& =6\left(\frac{1}{6 \sqrt{2}}+\frac{1}{3 \sqrt{2}}\right)\left(\frac{1}{3 \sqrt{2}}+\frac{2}{3 \sqrt{2}}\right) \\
& =6 \cdot \frac{1}{2} \cdot 1=3
\end{aligned}
\)

Hint

We have 10 arithmetic progressions (A.P.s) with the first terms \(a_i\) and the common differences \(d_i\), where \(i=1,2, \ldots, 10\).
The first terms are \(a_i=i\) and the common differences are \(d_i=2 i-1\).
Now, we need to find the sum of the first 12 terms for each A.P. The formula for the sum of the first \(n\) terms of an A.P. is:
\(
S_n=n\left(\frac{2 a+(n-1) d}{2}\right)
\)

In this case, we need to find the sum of the first 12 terms for each A.P., so we have:
\(
S_{12}=12\left(\frac{2 a+11 d}{2}\right)
\)
Now, we can compute the sum \(s_i\) for each A.P.:
\(
s_i=12\left(\frac{2 i+11(2 i-1)}{2}\right)=6(2 i+22 i-11)=6(24 i-11)
\)
Finally, we need to find the sum of all \(s_i\) for \(i=1,2, \ldots, 10\) :
\(
\sum_{i=1}^{10} s_i=6 \sum_{i=1}^{10}(24 i-11)=6\left(24 \sum_{i=1}^{10} i-11 \sum_{i=1}^{10} 1\right)
\)
The sum of the first 10 integers is \(\sum_{i=1}^{10} i=\frac{10(10+1)}{2}=55\), so we have:
\(
\sum_{i=1}^{10} s_i=6(24 \cdot 55-11 \cdot 10)=6(1320-110)=6 \cdot 1210=7260
\)
Thus, the sum \(\sum_{i=1}^{10} s_i\) is equal to 7260 .

Hint

Given the sum of the first \(n\) terms, \(S_n=\frac{n^2+3 n}{(n+1)(n+2)}\), we can find the \(n ^{\text {th }}\) term \(a_n\) as the difference between the sum of the first \(n\) terms and the sum of the first \(n-1\) terms :
So, \(a_n=S_n-S_{n-1}\)
Solving, we get :
\(
a_n=\frac{n^2+3 n}{(n+1)(n+2)}-\frac{(n-1)^2+3(n-1)}{n(n+1)}
\)
Simplifying further, we find :
\(
a_n=\frac{4}{n(n+1)(n+2)}
\)

Then, we find the reciprocal of \(a_n\) :
\(
\frac{1}{a_n}=\frac{n(n+1)(n+2)}{4}
\)
Now, we sum this over the first 10 terms :
\(
\sum_{k=1}^{10} \frac{1}{a_k}=\sum_{k=1}^{10} \frac{k(k+1)(k+2)}{4}
\)
Evaluating the sum :
\(
\sum_{k=1}^{10} \frac{1}{a_k}=\frac{1}{16}\left[\sum_{k=1}^{10} k(k+1)(k+2)(k+3)-(k-1) k(k+1)(k+2)\right]
\)
This can be rewritten as the sum of differences :
\(
\sum_{k=1}^{10} \frac{1}{a_k}=\frac{1}{16}[(1 \cdot 2 \cdot 3 \cdot 4-0)+(2 \cdot 3 \cdot 4 \cdot 5-1 \cdot 2 \cdot 3 \cdot 4)+\cdots+(10 \cdot 11 \cdot 12 \cdot 13-9 \cdot 10 \cdot 11 \cdot 12)]
\)
\(
\sum_{k=1}^{10} \frac{1}{a_k}=\frac{1}{16}(10 \cdot 11 \cdot 12 \cdot 13-0)=\frac{1}{16} \cdot 17160
\)
Now, given the condition that :
\(
28 \sum_{k=1}^{10} \frac{1}{a_k}=p_1 p_2 p_3 \ldots p_m
\)
Substituting the sum we’ve calculated:
\(28 \cdot \frac{1}{16} \cdot 17160=p_1 p_2 p_3 \ldots p_m\)
This simplifies to :
\(
30030=p_1 p_2 p_3 \ldots p_m
\)
The prime factorization of 30030 is \(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13\), which consists of 6 primes.
Therefore, \(m\) is equal to 6 .

Hint

Given that \(a+b+c+d=11\) and the maximum value of \(a^5 b^3 c^2 d\) is \(3750 \beta\), you assumed the numbers to be \(\frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{a}{5}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, \frac{c}{2}, \frac{c}{2}, d\).

Applying the AM-GM inequality:
\(
\frac{\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}+d}{11} \geq\left(\frac{\left(a^5 b^3 c^2 d\right)}{5^5 3^3 2^2 1}\right)^{\frac{1}{11}}
\)
Since \(a+b+c+d=11\), we have:
\(
1 \geq\left(\frac{\left(a^5 b^3 c^2 d\right)}{5^5 3^3 2^2 1}\right)^{\frac{1}{11}}
\)
Now, raising both sides to the power of 11 :
\(
1^{11} \geq \frac{a^5 b^3 c^2 d}{5^5 3^3 2^2 1}
\)

From the given information, we know that \(a^5 b^3 c^2 d \geq 3750 \beta\) :
\(
5^5 3^3 2^2 \geq 3750 \beta
\)
Now, we can solve for \(\beta\) :
\(
\beta \leq \frac{1}{3750} \cdot 5^5 3^3 2^2
\)
Since we are looking for the maximum value of \(\beta\), we take the equality case:
\(
\beta=\frac{1}{3750} \cdot 5^5 3^3 2^2
\)
Calculating the value, we find that:
\(
\beta=90
\)
So, the value of \(\beta\) is 90 .

Hint

We have, mean of \(x_1, x_2 \ldots \ldots x_{100}=200\)
Where, \(x_1, x_2 \ldots x_{100}\) are in AP with first term as 2 .
\(
\begin{aligned}
& \text { Mean }=200 \\
& =\frac{\sum_{i=1}^{100} x_i}{100}=200 \\
& \frac{100}{2} \times[2 \times 2+99 d]=20000 \\
& \Rightarrow 4+99 d=400 \\
& \Rightarrow 99 d=396 \\
& d=4
\end{aligned}
\)
Also,
\(
\begin{aligned}
y_i & =i\left(x_i-i\right) \\
& =i[2+(i-1) 4-i] \\
& =i[3 i-2] \\
& =3 i^2-2 i
\end{aligned}
\)
\(
\begin{aligned}
\text { Required mean } & =\frac{\sum_{i=1}^{100} y_i}{100} \\
& =\frac{1}{100}\left[\sum_{i=1}^{100}\left(3 i^2-2 i\right)\right] \\
& =\frac{1}{100}\left[\frac{3 \times 100 \times 101 \times 201}{6}-2 \times \frac{100 \times 101}{2}\right] \\
& =\frac{20301}{2}-101 \\
& =10049.50
\end{aligned}
\)

Hint

\(
\begin{aligned}
&S _n=4+11+21+24+50+\ldots+ T _n\\
&\begin{aligned}
& S _n=4+11+21+34++ T _{n-1}+ T _n \\
& -\quad-\quad-\quad-\quad- \quad-\quad-\quad- \\
& \hline 0=4+7+10+13+16+\ldots\left( T _n- T _{n-2}\right)- T _n
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow T _n=4+7+10+13+16+\ldots \text { to } n \text { terms } \\
& \Rightarrow T _n=\frac{n}{2}[2 \times 4+(n-1) 3] \\
& T _n=\frac{3}{2} n^2+\frac{5}{2} n
\end{aligned}
\)
So \(S _n=\Sigma T _n=\frac{3}{2} \Sigma n^2+\frac{5}{2} \Sigma n\)
\(
\begin{aligned}
& \Rightarrow S _n=\frac{3}{2} \times \frac{n(n+1)(2 n+1)}{6}+\frac{5}{2} \times \frac{n(n+1)}{2} \\
& \Rightarrow S_n=\frac{n(n+1)}{4}(2 n+1+5)=\frac{n(n+1)(n+3)}{2}
\end{aligned}
\)

Hence, \(\frac{1}{60}\left(S_{29}-S_9\right)=\frac{1}{60} \times \frac{1}{2}(29 \times 30 \times 32-9 \times 10 \times 12)\)
\(
=223
\)

Hint

Given that the first term \(a\) and common ratio \(r\) of a geometric progression be positive integer. So, their 1 st three terms are \(a, a r, a r^2\)
According to the question, \(a^2+a^2 r^2+a^2 r^4=33033\)
\(
\begin{aligned}
& \Rightarrow a^2\left(1+r^2+r^4\right)=3 \times 7 \times 11 \times 11 \times 13 \\
& =3 \times 7 \times 13 \times 11^2 \\
& \therefore \quad a^2=11^2 \\
& \Rightarrow \quad a=11 \\
&
\end{aligned}
\)
and \(1+r^2+r^4=273\)
\(
\begin{aligned}
& \Rightarrow r^2+r^4=272 \\
& \Rightarrow r^4+r^2-272=0 \\
& \Rightarrow\left(r^2+17\right)\left(r^2-16\right)=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow r^2=-17(\text { not possible) } \\
& \Rightarrow r^2-16=0 \\
& \Rightarrow r= \pm 4 \\
& \Rightarrow r=4(\because r>0)
\end{aligned}
\)
So, sum of these first three terms is \(a+a r+a r^2=11+44+176=231\)

Hint

Let \(S _n=5+8+14+23+\ldots+a_n\) and \(S _n=0+5+8+14+\ldots .+a_n\)
On subtracting, we get
\(
\begin{aligned}
& 0=5+3+6 \ldots-a_n \\
& \Rightarrow a_n=5+3+6+9+\ldots(n-1) \text { terms } \\
& =5+\left[\frac{(n-1)}{2}(6+(n-2) 3)\right] \\
& =5+\left[\frac{(n-1)}{2}(6+3 n-6)\right] \\
& =5+\frac{(n-1)(3 n)}{2} \\
& =\frac{10+3 n^2-3 n}{2}
\end{aligned}
\)
\(
\begin{aligned}
& \text { So, } a_{40}=\frac{3(40)^2-3(40)+10}{2} \\
& =\frac{4800-120+10}{2}=2345 \\
& \text { Now, } S_n=\sum_{k=1}^n a_k \\
& \Rightarrow S_{30}=\frac{3 \sum_{n=1}^{30} n^2-3 \sum_{n=1}^{30} n+10 \sum_{n=1}^{30} 1}{2} \\
& =\frac{3 \times(30)(30+1)(60+1)}{12}-\frac{3 \times 30 \times 31}{4}+\frac{10 \times 30}{2} \\
& =\frac{28365-1395+300}{2}=\frac{27270}{2} \\
& =13635 \\
& \therefore S_{30}-a_{40}=13635-2345=11290
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \because S_k=\frac{1+2+\ldots+k}{k} \\
& =\frac{k(k+1)}{2 k}=\frac{k+1}{2} \\
& \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\
& \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\right] \\
& =\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+n+\frac{2 n(n+1)}{2}\right] \\
& =\frac{n}{4}\left[\frac{(n+1)(2 n+1)}{6}+1+n+1\right] \\
& =\frac{n}{24}\left[2 n^2+3 n+1+6+6 n+6\right] \\
& =\frac{n}{24}\left[2 n^2+9 n+13\right]
\end{aligned}
\)
On comparing, we get
\(
A =24, B =2, C =9, D =13
\)
(A) \(A+B+C+D=48\), which is not divisible by 5 .
(B) \(A + C + D =46\), which is divisible by 2.
(C) \(A+B=26\)
\(
\begin{aligned}
& 5( D – C )=5(13-9)=20 \\
& \therefore 26 \neq 20
\end{aligned}
\)
(D) \(A+B=24+2=26\), divisible by 13 .

Hint

\(
\begin{aligned}
& \text { Given } \operatorname{gcd}(m, n)=1 \text { and } \\
& \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\
& =1012 m^2 n \\
& \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\
& =1012 m^2 n \\
& \Rightarrow(1-2)(1+2)+(3-4)(3+4)+\ldots .(2021-2022) \\
& (2021+2022)+(2023)^2=(1012) m^2 n \\
& \Rightarrow(-1)(1+2)+(-1)(3+4)+\ldots .+ \\
& (-1)(2021+2022)+\left(2023^2\right)=(1012) m^2 n \\
& \Rightarrow(-1)[1+2+3+4+\ldots+2022]+(2023)^2=1012 m^2 n \\
& \Rightarrow(-1)\left[\frac{(2022) \cdot(2022+1)}{2}\right]+(2023)^2=(1012) m^2 n \\
& \Rightarrow(-1)\left[\frac{(2022)(2023)}{2}\right]+(2023)^2=(1012) m^2 n \\
&
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow(2023)[2023-1011]=(1012) m^2 n \\
& \Rightarrow m^2 n=2023 \\
& \Rightarrow m^2 n=289 \times 7 \\
& \Rightarrow m^2=289 \text { and } n=7 \\
& \Rightarrow m=17 \text { and } n=7(\text { such that } \operatorname{gcd}(m, n)=1) \\
& \therefore m^2-n^2=289-49=240
\end{aligned}
\)

Hint

\(
\begin{aligned}
& S _n=5+11+19+29+41+\ldots+ T _n \\
& S _n=\quad 5+11+19+29+\ldots .+ T _{n-1}+ T _n \\
& \hline 0=5+6+8+10+12+\ldots \ldots T _n
\end{aligned}
\)
\(
\begin{aligned}
& 0=5+\frac{n-1}{2}[2 \times 6+(n-2)(2)]-T_n \\
& \Rightarrow T_n=5+(n-1)(n+4) \\
& \Rightarrow T_n=5+n^2+3 n-4 \\
& \Rightarrow T_n=n^2+3 n+1 \\
& \Sigma T_n=\Sigma n^2+3 \Sigma n+\Sigma 1 \\
& \Rightarrow \quad S_n=\frac{n(n+1)(2 n+1)}{6}+\frac{3 n(n+1)}{2}+n
\end{aligned}
\)
When, \(n=20\)
Then,
\(
\begin{aligned}
S_{20} & =\frac{20 \times 21 \times 41}{6}+\frac{3 \times 20 \times 21}{2}+20 \\
& =2870+630+20=3520
\end{aligned}
\)