Question

The sum of first three terms of a G.P. is $16$ and the sum of 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 the G.P. be $a,ar,{ar}^2,{ar}^3,...$
According to the given condition, $a+ar+{ar}^2=16$
and ${ar}^3+{ar}^4+{ar}^5=128$
$\Rightarrow a(1+r+r^2)=16....\left(1\right)$
& ${ar}^3(1+r+r^2)=128...\left(2\right)$
Dividing equation (2) by (1) , we obtain
$\frac{{ar}^3(1+r+r^2)}{a(1+r+r^2)}=\frac{128}{16}\Rightarrow r^3=8\therefore r=2$
Substituting $r=2$ in (1) ,
we obtain $a(1+2+4)=16\Rightarrow a\left(7\right)=16\therefore a=\frac{16}{7}$
And sum of first $n$ terms of this G.P. Is given by,
$S_n=\frac{a(r^n-1)}{r-1}=\frac{16/7(2^n-1)}{2-1}=\frac{16}{7}(2^n-1)$