The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is

#### Options

(a) 1/2

(b) 2/3

(c) 1/3

(d) −1/2

#### Solution

(a) 1/2

\[\text{ Let the G . P . be a, ar }, a r^2 , a r^3 , . . . , \infty . \]

\[ S_\infty = 4\]

\[ \Rightarrow \frac{a}{1 - r} = 4 (i)\]

\[\text{ Also, sum of the cubes }, S_1 = 92\]

\[ \Rightarrow \frac{a^3}{\left( 1 - r^3 \right)} = 92 (ii)\]

\[\text{ Putting the value of a from } (i) \text{ to } (ii): \]

\[ \Rightarrow \frac{\left( 4(1 - r) \right)^3}{\left( 1 - r^3 \right)} = 92\]

\[ \Rightarrow \frac{64(1 - r )^3}{\left( 1 - r^3 \right)} = 92\]

\[ \Rightarrow \frac{\left( 1 - r \right)^3}{\left( 1 - r \right)\left( 1 + r + r^2 \right)} = \frac{92}{64}\]

\[ \Rightarrow \frac{\left( 1 - r \right)^2}{\left( 1 + r + r^2 \right)} = \frac{23}{16}\]

\[ \Rightarrow 16\left( 1 - 2r + r^2 \right) = 23\left( 1 + r + r^2 \right)\]

\[ \Rightarrow 7 r^2 + 55r + 7 = 0\]

\[\text{ Using the quadratic formula }: \]

\[ \Rightarrow r = \frac{- 55 + \sqrt{{55}^2 - 4 \times 7 \times 7}}{2 \times 7}\]

\[ \Rightarrow r = \frac{- 55 + \sqrt{{55}^2 - {14}^2}}{14}\]

\[ \Rightarrow r = \frac{- 55 + \sqrt{2829}}{14}\]