Sum of an infinite G.P. is 2 and sum of their cubes is 24, then 5th term of the G.P. is

A
3/16
B
3/8
C
– 3/8
D
– 3/16

Solution:

$\begin{array}{l} \frac{a}{1 - r} = 2, \frac{a^{3}}{1 - r^{3}} = 24 \\ ∴ \frac{8 (1 - r)^{3}}{1 - r^{3}} = 24 \Rightarrow \frac{1 - 2 r + r^{2}}{1 + r + r^{2}} = 3 \\ \Rightarrow 2 r^{2} + 5 r + 2 = 0 \Rightarrow r = - 2, - 1 / 2 \end{array}$

As $| r | < 1, r = - 1 / 2, a = 3$

Now, $t_{5} = a r^{4} = 3 / 16$

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