Quantitative Reasoning

Miscellaneous

Hard

The sum of an Infinite geometric series is 4 and the sum of the cubes of the terms of the same GP is 192. The common ratio of the original geometric series is:

$\frac{1}{2}$

$-\frac{1}{2}$

$\frac{1}{4}$

$-\frac{1}{4}$

✅ Correct Option: 2

Let first term $a$ and ratio $r$ . Then $a/(1-r) = 4$ , so $a = 4(1-r)$ . Sum of cubes: $a^3/(1-r^3) = 192$ . Substituting: $64(1-r)^3/(1-r^3) = 192$ . Using $1-r^3 = (1-r)(1+r+r^2)$ : $64(1-r)^2/(1+r+r^2) = 192$ . This gives $2r^2 + 5r + 2 = 0$ , so $r = -1/2$ or $-2$ . Since $|r| < 1$ , $r = -1/2$ .

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The sum of an Infinite geometric series is 4 and the sum of the cubes of the terms of the same GP is 192. The common ratio of the original geometric series is:

Solution

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