Question

1
If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first for term

Answer 1

If the first term is $a$, then the second term is $ar$, the third is $ar^2$, the fourth is $ar^3$, and the fifth is $ar^4$.

We're given

$\begin{cases}ar=6\\ar^4=48\end{cases}\implies\dfrac{ar^4}{ar}=r^3=8\implies r=2\implies a=3$

So the first five terms in the GP are

3, 6, 12, 24, 48

Adding up the first four gives a sum of 45.

If you were asked to find the sum of many, many more terms, having a formula for the n-th partial sum would convenient. Let $S_n$ denote the sum of the first n terms in the GP:

$S_n=3+3\cdot2+3\cdot2^2+\cdots+3\cdot2^{n-2}+3\cdot2^{n-1}$

Multiply both sides by 2:

$2S_n=3\cdot2+3\cdot2^2+3\cdot2^3+\cdots+3\cdot2^{n-1}+3\cdot2^n$

Subtract this from $S_n$, which eliminates all the middle terms:

$S_n-2S_n=3-3\cdot2^n\implies -S_n=3(1-2^n)\implies S_n=3(2^n-1)$

Then the sum of the first four terms is again $S_4=3(2^4-1)=45$.