The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Find the sum of n terms of the G.P.

Question

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Find the sum of $n$ terms of the G.P.

Smart Achievers · Accepted Answer

(1)

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

$Then, a + a r + a r^{2} = 16 \Rightarrow a (1 + r + r^{2}) = 16$ ...(i)

$and a r^{3} + a r^{4} + a r^{5} = 128 \Rightarrow a r^{3} (1 + r + r^{2}) = 128$ ...(ii)

$Dividing (ii) by (i), we get$

$\frac{a r^{3} (1 + r + r^{2})}{a (1 + r + r^{2})} = \frac{128}{16} \Rightarrow r^{3} = 8 \Rightarrow r = 2$

$Putting r = 2 in (i), we get a = \frac{16}{7}$

$∴$ $S_{n} = a (\frac{r^{n} - 1}{r - 1}) \Rightarrow S_{n} = \frac{16}{7} (\frac{2^{n} - 1}{2 - 1}) = \frac{16}{7} (2^{n} - 1)$

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