Question

The sum of an infinitely decreasing geometric progression is $1.5$ and the sum of the squares of its terms is $\frac{1}{8}$. Find the progression.

Answer 1

$\frac{a}{1-r}=1.5=\frac{15}{10}=\frac{3}{2}$

$\Rightarrow \frac{a^2}{1-r^2}=\frac{1}{8}$

Now, $8a^2=1-r^2$

$\Rightarrow 8\times \left[\frac{3}{2}\right(1-r){]}^2=1-r^2$

$\Rightarrow 8\times \frac{9}{4}(1-r{)}^2=1-r^2$

$\Rightarrow 18(1+r^2-2r)=1-r^2$

$\Rightarrow 18+18r^2-36r=1-r^2$

$\Rightarrow 19r^2-36r+17=0$