Question

Progression.
The sum of an infinitely decreasing geometric progression is 12 and the sum of the squares of its terms is 48. Find the sum of the first ten terms of the progression.

Answer 1

Let the first term of GP is $a$ and $r$ be the common ratio.

$\frac{a}{1-r}=12$

And for the GP of square of terms, the first term will be $a^2$ and the common ratio would be $r^2$-

$\frac{a^2}{1-r^2}=48$

Now squaring first equation and dividing it by second equation we get-

$\frac{1-r^2}{(1-r{)}^2}=\frac{{12}^2}{48}$

$\frac{(1-r)(1+r)}{(1-r{)}^2}=3$

$⟹\frac{1+r}{1-r}=3$

$⟹1+r=3-3r$

$⟹r=\frac{1}{2}$

Again-

$\frac{a}{1-r}=12$

$⟹a=12\times \frac{1}{2}=6$

Sum of first ten terms-

$S_{10}=\frac{a(r^{10}-1)}{r-1}$

$=\frac{6(1-\frac{1}{2^{10}})}{\frac{1}{2}}$

$=12(1-\frac{1}{2^{10}})$