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