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:

$-\frac{1}{2}$

The sum of an Infinite geometric series = 4

a/(1-r) = 4 ........(1)

cubing both sides of equation (1)

a³=64(1−r)³

Put value of a3 in equation 2

192(1−r³) = 64(1−r)³

64(1−r)³=192(1−r)(1+r²+r)

(1−r)²=3(1+r²+r)

(1−r)²=3(1+r²+r)

2r²+5r+2=0

Solving for r

r = -1/2

The correct answer is Option (2) → $-\frac{1}{2}$