The sum of the first four terms of a geometric progression (G.P) is $30$ and the sum of the next four terms is $480$. Find the sum of the first twelve terms.
Let the first term of the G.P be $= t_1 = a$
and the common ratio be $= r$
Sum of $n$ terms of G.P $= S_n = \dfrac{a \left(r^n - 1\right)}{r - 1}$
Given: $\;$ Sum of first four terms of G.P $= S_4 = 30$
i.e. $\;$ $S_4 = \dfrac{a \left(r^4 - 1\right)}{r - 1} = 30$ $\;\;\; \cdots \; (1)$
And: $\;$ Sum of next four terms $= 480$
i.e. $\;$ Sum of first $8$ terms $-$ Sum of first $4$ terms $= 480$
i.e. $\;$ $S_8 - S_4 = 480$
i.e. $\;$ $S_8 = 480 + S_4 = 480 + 30 = 510$ $\;\;\; \left[\because \; S_4 = 30\right]$
i.e. $\;$ $S_8 = \dfrac{a \left(r^8 - 1\right)}{r - 1} = 510$ $\;\;\; \cdots \; (2)$
$\therefore \;$ We have from equations $(1)$ and $(2)$,
$\dfrac{S_8}{S_4} = \dfrac{r^8 - 1}{r^4 - 1} = \dfrac{510}{30} = 17$
i.e. $\;$ $\dfrac{\left(r^4\right)^2 - 1^2}{r^4 - 1} = 17$
i.e. $\;$ $\dfrac{\left(r^4 + 1\right) \left(r^4 - 1\right)}{r^4 - 1} = 17$
i.e. $\;$ $r^4 + 1 = 17$
i.e. $\;$ $r^4 = 16 = 2^4$ $\implies$ $r = 2$
Substituting the value of $r$ in equation $(1)$ gives
$30 = \dfrac{a \left(2^4 - 1\right)}{2 - 1}$
i.e. $\;$ $30 = 15 \times a$ $\implies$ $a = 2$
$\therefore \;$ Sum of first twelve terms of G.P
$= S_{12} = \dfrac{2 \times \left(2^{12} - 1\right)}{2 - 1} = 2 \times 4095 = 8190$