9.7 Relationship Between A.M. and G.M.

Let \(A\) and \(G\) be A.M. and G.M. of two given positive real numbers \(a\) and \(b\), respectively. Then
\(
A =\frac{a+b}{2} \text { and } G =\sqrt{a b}
\)
Thus, we have
\(
\begin{aligned}
A – G & =\frac{a+b}{2}-\sqrt{a b}=\frac{a+b-2 \sqrt{a b}}{2} \\
& =\frac{(\sqrt{a}-\sqrt{b})^2}{2} \geq 0 \dots(1)
\end{aligned}
\)
From (1), we obtain the relationship \(A \geq G\).

Example 1: If A.M. and G.M. of two positive numbers \(a\) and \(b\) are 10 and 8 , respectively, find the numbers.

Solution: Given that
A.M. \(=\frac{a+b}{2}=10 \dots(1)\)
and \(\text { G.M. }=\sqrt{a b}=8 \dots(2)\)
From (1) and (2), we get
\(
\begin{aligned}
& a+b=20 \dots(3) \\
& a b=64 \dots(4)
\end{aligned}
\)
Putting the value of \(a\) and \(b\) from (3), (4) in the identity \((a-b)^2=(a+b)^2-4 a b\), we get
\(
(a-b)^2=400-256=144
\)
or \(a-b= \pm 12\)
Solving (3) and (5), we obtain
\(
a=4, b=16 \text { or } a=16, b=4
\)
Thus, the numbers \(a\) and \(b\) are 4,16 or 16,4 respectively.

Example 2: If \(p^{\text {th }}, q^{\text {th }}, r^{\text {th }}\) and \(s^{\text {th }}\) terms of an A.P. are in G.P, then show that \((p-q),(q-r),(r-s)\) are also in GP.

Solution: Here
\(
\begin{aligned}
& a_p=a+(p-1) d \dots(1) \\
& a_q=a+(q-1) d \dots(2) \\
& a_r=a+(r-1) d \dots(3) \\
& a_s=a+(s-1) d \dots(4)
\end{aligned}
\)
Given that \(a_p, a_q, a_r\) and \(a_s\) are in G.P.
So \(\frac{a_q}{a_p}=\frac{a_r}{a_q}=\frac{a_q-a_r}{a_p-a_q}=\frac{q-r}{p-q} \text { (why ?) } \dots(5)\)
Similarly
\(
\frac{a_r}{a_q}=\frac{a_s}{a_r}=\frac{a_r-a_s}{a_q-a_r}=\frac{r-s}{q-r} \text { (why ?) } \dots(6)
\)
Hence, by (5) and (6)
\(\frac{q-r}{p-q}=\frac{r-s}{q-r}\), i.e., \(p-q, q-r\) and \(r-s\) are in G.P.

Example 3: If \(a, b, c\) are in G.P. and \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}\), prove that \(x, y, z\) are in A.P.

Solution: Let \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k\) Then
\(
a=k^x, b=k^y \text { and } c =k^z \text {. } \dots(1)
\)
Since \(a, b, c\) are in G.P., therefore,
\(
b^2=a c \dots(2)
\)
Using (1) in (2), we get
\(
k^{2 y}=k^{x+z} \text {, which gives } 2 y=x+z
\)
Hence, \(x, y\) and \(z\) are in A.P.

Example 4: If \(a, b, c, d\) and \(p\) are different real numbers such that \(\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p+\left(b^2+c^2+d^2\right) \leq 0\), then show that \(a, b, c\) and \(d\) are in GP.

Solution: Given that
\(
\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p+\left(b^2+c^2+d^2\right) \leq 0 \dots(1)
\)
But L.H.S.
\(
=\left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2-2 b c p+c^2\right)+\left(c^2 p^2-2 c d p+d^2\right) \text {, }
\)
which gives \((a p-b)^2+(b p-c)^2+(c p-d)^2 \geq 0 \dots(2)\)
Since the sum of squares of real numbers is non negative, therefore, from (1) and (2), we have, \(\quad(a p-b)^2+(b p-c)^2+(c p-d)^2=0\)
or \(a p-b=0, b p-c=0, c p-d=0\)
This implies that \(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p\)
Hence \(a, b, c\) and \(d\) are in GP.

Example 5: If \(p, q, r\) are in G.P. and the equations, \(p x^2+2 q x+r=0\) and \(d x^2+2 e x+f=0\) have a common root, then show that \(\frac{d}{p}, \frac{e}{q}, \frac{f}{r}\) are in A.P.

Solution: The equation \(p x^2+2 q x+r=0\) has roots given by
\(
x=\frac{-2 q \pm \sqrt{4 q^2-4 r p}}{2 p}
\)
Since \(p, q, r\) are in G.P. \(q^2=p r\). Thus \(x=\frac{-q}{p}\) but \(\frac{-q}{p}\) is also root of \(d x^2+2 e x+f=0\) (Why ?). Therefore
\(
d\left(\frac{-q}{p}\right)^2+2 e\left(\frac{-q}{p}\right)+f=0,
\)
or \(\quad d q^2-2 e q p+f p^2=0 \dots(1)\)
Dividing (1) by \(p q^2\) and using \(q^2=p r\), we get
\(
\frac{d}{p}-\frac{2 e}{q}+\frac{f p}{p r}=0, \text { or } \quad \frac{2 e}{q}=\frac{d}{p}+\frac{f}{r}
\)
Hence \(\frac{d}{p}, \frac{e}{q}, \frac{f}{r}\) are in A.P.