The sum of an infinite number of terms of GP is 4 and the sum of their cubes is 192 - Maths - Sequences and Series

Question

The sum of an infinite number of terms of GP is 4 and the sum of
their cubes is 192 Find the series

Aakash Sharma · Accepted Answer

Let a be the first term and r be the common ratio of the given G.P. then,

Sum = 4

$\Rightarrow \frac{a}{1 - r} = 4 ........... (1)$

Sum of cubes = 192

⇒ a³ + a³r³ + a³r⁶+ ......... = 192

$\Rightarrow \frac{a^{3}}{1 - r^{3}} = 192 ........... (2)$

Dividing the cube of (1) by (2) we get

$\frac{a^{3}}{1 - r^{3}} \cdot \frac{1 - r^{3}}{a^{3}} = \frac{(4)^{3}}{192} \Rightarrow \frac{1 - r^{3}}{(1 - r)^{3}} = \frac{1}{3} \Rightarrow \frac{1 + r + r^{2}}{(1 - r)^{2}} = \frac{1}{3}$

⇒ 2r² + 5r + 2 = 0

⇒ (2r + 1) (r + 2) = 0

$\Rightarrow r = \frac{- 1}{2} or r = - 2$

Since r ≠ – 2 because –1 < r < 1 for an infinite G.P.

$Thus r = \frac{- 1}{2} Putting r = \frac{- 1}{2} th (1) we get \frac{a}{1 - (\frac{- 1}{2})} = 4 \Rightarrow a = 6$

Hence the G.P. is $6, - 3, \frac{3}{2} ........$

The sum of an infinite number of terms of GP is 4 and the sum of their cubes is 192. Find the series.