First slide

Geometric progression

Question

$If the sum of an infinite decreasing G.P is 3 and the sum of the squares of its$ $terms is \frac{9}{2} then the sum of the cubes of the terms is$

Moderate

A

$\frac{1}{13}$
B

$\frac{100}{13}$
C

$\frac{101}{13}$
D

$\frac{108}{13}$

Solution

${Let the G.P terms are a, ar, ar}^{2}, \dots where 0 < r < 1$
$\begin{array}{l} {Given a+ar+ar}^{2} + \dots = 3, and a^{2} + a^{2} r^{2} + a^{2} r^{4} + \dots = 9 / 2 \\ \to \frac{a}{1 - r} = 3 - - - - (1) \\ a n d \frac{a^{2}}{1 - r^{2}} = \frac{9}{2} - - - (2) \\ \Rightarrow r = 1 / 3, a = 2 \\ The sum of the cubes \Rightarrow a^{3} + a^{3} r^{3} + a^{3} r^{6} + \dots \infty \\ = \frac{a^{3}}{1 - r^{3}} = \frac{8}{1 - 1 / 27} = \frac{108}{13} \end{array}$

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