For a sequence, if $S_{n} = 3 (2^{n} - 1)$ , find $n^{t h}$ term, hence show that it is a G.P. Also find the common ratio.

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Solution

$S_{n} = 3 (2^{n} - 1)$
$S_{n - 1} = 3 (2^{n - 1} - 1)$
$T_{n} = S_{n} - S_{n - 1}$
$= 3 \cdot 2^{n} - 3 - {3.2}^{n - 1} + 3$
$T_{n} = {3.2}^{n - 1} (2 - 1) = {3.2}^{n - 1}$
$∴ T_{1} = 3. (2)^{1 - 1} = 3$
$T_{2} = 3. (2)^{2 - 1} = 6$
$T_{3} = 3. (2)^{3 - 1} = 12$
Ratio. $d = \frac{T_{1}}{T_{2}} = \frac{3}{6} = \frac{1}{2}$
$d = \frac{T_{2}}{T_{3}} = \frac{6}{12} = \frac{1}{2}$
$∴$ Hence G.P.

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