The sum of an infinite G.P. is 2 and the sum of G.P. made from the cubes of this infinite
G.P. is 24. Find the G.P.

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3, \frac{3}{2}, \frac{3}{4}, \dots \dots \dots

4, \frac{3}{2}, \frac{3}{4}, \dots \dots \dots

3, - \frac{3}{2}, \frac{3}{4}, \dots \dots \dots

2, - \frac{2}{2}, \frac{2}{4}, \dots \dots \dots

answer is C.

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Detailed Solution

Given,
The sum of infinite geometric progression (GP)= 2
Now the sum of infinite GP,

S_{\infty} = \frac{a}{1 - r}

So,

\frac{a}{1 - r} = 2

….. (i)
The sum of the G.P. generated from this infinite G.P.'s cubes is 24.
Therefore,

\frac{a^{3}}{1 - r^{3}} = 24

\frac{a^{3}}{(1 - r)^{3}} \times \frac{1 - r^{3}}{a^{3}} = \frac{2^{3}}{24}

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

\frac{1 + r + r^{2}}{(1 - r)^{2}} = \frac{1}{3}

2 r^{2} + 5 r + 2 = 0

r = - \frac{1}{2}, - 2

For GP, we cannot take the common ration -2. So, take r=-1/2,
Putting the value of r in eq. (i),

\frac{a}{1 - r} = 2 \frac{a}{1 - (- \frac{1}{2})} = 2 a = 3

Thus, the first term is 3.

r = - \frac{1}{2}, a = 3

So, the required G.P. is

3, - \frac{3}{2}, \frac{3}{4}, \dots \dots \dots

Correct option is 3.

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