Mark the correct alternative in each of the following:

The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 192. The common ratio of the original G.P. is


Let, the first term of G.P. is a and common ratio is r.


We know that common ratio of infinite G.P. is belongs to


[0, 1)


G.P. a, ar, ar2, ……


Sum of infinite terms of G.P. = = 4


a = 4(1 – r)


Cubic terms of a G.P. a3, a3r3, a3r6, ……


Sum of cubes of terms = = 192


a3 = 192(1 – r3)


43(1 – r)3 = 92(1 – r3)


(1 – r)3 = 3(1 – r)(1 + r + r2)


Case I : 1 – r = 0


r = 1 (not possible)


Case II : (1 – r)2 = 3(1 + r + r2)


2r2 + 5r + 2 = 0


(2r + 1)(r + 2) = 0


r = -2 (not possible) and r = -1/2


So, common ratio of original G.P. is -1/2

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