Question

The sum of an infinite geometric series is 2 and the sum of the geometric series made from the cubes of this infinite series is 24. Then the series is –

• A. 3 + $\frac{3}{2}$ – $\frac{3}{4}$ + $\frac{3}{8}$ ….
• B. 3 + $\frac{3}{2}$ + $\frac{3}{4}$ + $\frac{3}{8}$ + ….
• C. 3 – $\frac{3}{2}$ + $\frac{3}{4}$ – $\frac{3}{8}$ + ….
• D. None of these

Answer 1

Answer: (C)

3 – $\frac{3}{2}$ + $\frac{3}{4}$ – $\frac{3}{8}$ + ….

Answer 2

Let first term = a, common ratio = r,

Where –1 < r < 1

Then, $\frac{a}{1 - r}$ = 2 and $\frac{a^{3}}{1 - r^{3}}$ = 24

\ $\frac{1 - r^{3}}{(1 - r)^{3}}$ = $\frac{1}{3}$

i.e. 1 – 2r + r² = 3 (1 + r + r²)

or 2r² + 5r + 2 = 0.

\ r = –2 or –$\frac{1}{2}$.

As –1 < r < 1 \ we have r = –$\frac{1}{2}$

Putting this value of r, we get a = 3,

\ The series is 3 – $\frac{3}{2}$ + $\frac{3}{4}$ – $\frac{3}{8}$ + ……

Hence (3) is the correct answer.