Question

Find $S_4:S_8$ for the series 4 + 12 + 36 + .............

Answer 1

It is given that the first term of G.P is $a_1=4$ and the second term is $a_2=12$, therefore,

$r=\frac{a_2}{a_1}=\frac{12}{4}=3$

We know that the formula for sum of $n$ terms of G.P is $S_n=\frac{a(r^n-1)}{r-1}$, where $a$ is the first term, $r$ is the common ratio.

In the given geometric series, the first term is $a=4$, the common ratio is $r=3$, therefore, consider the ratio of the sum $S_4$ and $S_8$ as follows:

$\frac{S_4}{S_8}=\frac{\frac{4(3^4-1)}{3-1}}{\frac{4(3^8-1)}{3-1}}=\frac{3^4-1}{3^8-1}=\frac{81-1}{6561-1}=\frac{80}{6560}=\frac{1}{82}$

Hence, $S_4:S_8=1:82$.