The sum of an infinite G.P. is 4 and the sum of the cubes of its terms

1.	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. isA. 1/2B. 2/3C. 1/3D. −1/2
Answer» 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. = a/(1-r) = 4 ⇒ a = 4(1 – r) Cubic terms of a G.P. ⇒ a³, a³r³, a³r⁶, …… Sum of cubes of terms = \(\frac{a^3}{1-r^3}\) = 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

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. isA. 1/2B. 2/3C. 1/3D. −1/2

Answer»

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. = a/(1-r) = 4

⇒ a = 4(1 – r)

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

Sum of cubes of terms = \(\frac{a^3}{1-r^3}\) = 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