Question

In a G.P. sum of n terms is 255, the last term is 128 and the common ratio is 2. Find n.

Answer 1

Given $T_n=a{r}^{n-1}=128,r=2$
$S_n=\frac{a(r^n-1)}{r-1}=255or\frac{a{r}^{n-1}.r-a}{r-1}=255or\frac{128.2-a}{2-1}=255\therefore 256-255=aora=1$.
Now $a{r}^{n-1}=128\therefore 1.(2{)}^{n-1}=2^7\therefore n-1=7orn=8.$