Question

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Answer 1

Let 'a' be the first term and 'r' be the common ratio of given G.P.

Then $a+ar+a{r}^2=16$

$\Rightarrow a(1+r+r^2)=16$ .......(1)

and $a{r}^3+a{r}^4+a{r}^5=128$

$\Rightarrow a{r}^3(1+r+r^2)=128$ .......(2)

$\therefore$ Putting the value $a(1+r+r^2)$ from (1) into (2), we have :

$16r^3=128$

$\Rightarrow r^3=\frac{128}{16}=8\Rightarrow r=2$

Now, from (1), we have

$a(1+2+2^2)=16\Rightarrow a-\frac{16}{7}$

We know that

$S_n=\frac{a(r^n-1)}{r-1}$ When r > 1

$\therefore S_n=\frac{\frac{16}{7}[2^n-1]}{(2-1)}=\frac{16}{7}(2^n-1)$