Question

Three numbers are in GP, whose sum is 13 and the sum of whose squares is 91. Find the numbers.

Answer 1

Let the numbers in GP be a, ar and $a{r}^2.$

Given, sum of numbers = $13\Rightarrow a+ar+a{r}^2=13$

$\Rightarrow a(1+r+r^2)=13...\left(i\right)$

and sum of the squares = 91

$\Rightarrow a^2+a^2{r}^2+a^2{r}^4=91$

$\Rightarrow a^2(1+r^2+r^4)=91$

$\Rightarrow a^2(1-r+r^2)(1+r+r^2)=91[\because 1+r^2+r^4=(1-r+r^2\left)\right(1+r+r^2\left)\right]$

$\Rightarrow a(1-r+r^2)a(1+r+r^2)=91$

$\Rightarrow a(1-r+r^2)\times 13=91$ [from Eq. (i)]

$\Rightarrow a(1-r+r^2)=\mathrm{7...}\left(ii\right)$

On dividing Eq. (i) by Eq. (ii), we get

$\frac{1+r+r^2}{1-r+r^2}=\frac{13}{7}$

$\Rightarrow 7+7r+7r^2=13-13r+13r^2$

$\Rightarrow 6r^2-20r+6=0$

$\Rightarrow 3r^2-10r+3=0$

$\Rightarrow (3r-1)(r-3)=0$

$\Rightarrow r=3orr=\frac{1}{3}$

when r = 3, then from Eq. (i), we have

$a(1+3+3^2)=13\Rightarrow a=1$

and when $r=\frac{1}{3}$, then from Eq. (i), we have

$a(1+\frac{1}{3}+\frac{1}{3^2})=13\Rightarrow a=9$

So, the numbers are 1, 3 and 9 or 9, 3 and 1.