Question

The sum of three numbers which form a geometric progression is $13$ and the sum of their squares is $91$. Find the numbers.

Answer 1

Let the three terms be $a,ar,a{r}^2$

From the given conditions

$a+ar+a{r}^2=\mathrm{13.....}\left(1\right)$

$a^2+a^2{r}^2+a^2{r}^4=\mathrm{91.....}\left(2\right)$

$(1{)}^2-(2)⟹{13}^2-91=2(a^{2}r\left)\right(1+r+r^2)⟹ar=3$

$⟹\frac{3}{r}+3+3r=13⟹\frac{3+3r^2}{r}=10⟹r=3,\frac{1}{3}$

The three numbers are $1,3,9$