Question

The sum of the first four terms of a geometric progression is $30$ and the sum of the next four terms is $480$. Find the sum of the first twelve terms.

Answer 1

Sum of $1^{st}$ $4$ terms$=30$

$\Rightarrow \frac{a_1(r^4-1)}{r-1}=30-\left(i\right)$

Sum of $1^{st}$ $8$ terms$=30+480$

$\Rightarrow \frac{a_1(r^8-1)}{r-1}=510\Rightarrow \frac{a_1(r^4-1)(r^4+1)}{r-1}=510-\left(ii\right)$

Dividing $\left(ii\right)$ by $\left(i\right)$ we get

$\frac{\frac{a_1(r^4-1)(r^4+1)(r-1)}{r-1}}{\frac{a_1(r^4-1)}{r-1}}=\frac{510}{30}\Rightarrow \frac{a_1(r^4-1)(r^4+1)(r-1)}{a_1(r^4-1)(r-1)}=\frac{510}{30}\Rightarrow r^4+1=17\Rightarrow r^4=16\Rightarrow r=2$

$\therefore$ From $\left(i\right)$ replacing $r=2$

$\Rightarrow \frac{a_1(2^4-1)}{2-1}=30\Rightarrow a_1\times (16-1)=30\times 1\Rightarrow a_1\times 15=30\Rightarrow a_1=2$

$\therefore$ Sum of first $12$ terms

$S_{12}=\frac{a_1(r^{12}-1)}{r-1}=\frac{2\times (2^{12}-1)}{2-1}=2\times (2^{12}-1)=2\times 4095=8190$