If the sum of an infinite decreasing G.P. is 3 and sum of the cubes of its terms is 108/13, then common ratio is given by

A
$\frac{1}{5}$
B
$\frac{1}{4}$
C
$\frac{1}{3}$
D
$\frac{1}{2}$

Solution:

$\frac{a}{1 - r} = 3, \frac{a^{3}}{1 - r^{3}} = \frac{108}{13}$ where $| r | < 1$

$∴ \frac{27 (1 - r)^{3}}{1 - r^{3}} = \frac{108}{13}$

$\begin{array}{l} \Rightarrow \frac{1 - 2 r + r^{2}}{1 + r + r^{2}} = \frac{4}{13} \\ \Rightarrow 13 - 26 r + 13 r^{2} = 4 + 4 r + 4 r^{2} \\ \Rightarrow 9 r^{2} - 30 r + 9 = 0 \Rightarrow 3 r^{2} - 10 r + 3 = 0 . \\ \Rightarrow (3 r - 1) (r - 3) = 0 \Rightarrow r = 1 / 3 [∵ | r | < 1] \end{array}$

Solution:

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