Question

The sum of the first three terms of a G.P. is $13$ and sum of their squares is $91$. Determine the G.P.

Answer 1

Let the three terms be $\frac{a}{r},a,ar$

We are given $\frac{a}{r}+a+ar=13$ ....(i) and ${\left(\frac{a}{r}\right)}^2+a^2+{ar}^2=91$ ....(ii)

Squaring equation (i) and substituting (ii), we get

$91+2\times \frac{a^2}{r}+2a^2+2a^{2}r=169$

$\Rightarrow 2a^2(\frac{1}{r}+1+r)=78$

Divide by equation (i)

We get $a=3$

Solve in (i)

We get $r=3,\frac{1}{3}$

Therefore corresponding G.P

$1,3,\mathrm{9...}$ and $9,3,\mathrm{1....}$