sum of an infinite gp is 57 and sum of their cube is 9747 find the gp - Maths - Sequences and Series

Question

sum of an infinite gp is 57 and sum of their cube is 9747 find
the gp

Ajanta Trivedi · Accepted Answer

let the first term of the GP be a and the common ratio be r where -1<r<1

the GP is $a, a r, a r^{2}, ........$

therefore the sum of the infinite terms of the GP is

$\frac{a}{1 - r} = 57.......... (1)$

if taking the cube of each terms the new GP is $a^{3}, a^{3} r^{3}, a^{3} r^{6}, ............$

therefore the sum of their cube is

$\frac{a^{3}}{1 - r^{3}} = 9747 ............ (2)$

taking the cube of the (1) : $\frac{a^{3}}{(1 - r)^{3}} = 57^{3} \Rightarrow a^{3} = (57)^{3} \cdot (1 - r)^{3}$ ........(3)

substituting the value of a in terms of r in (2),

$\frac{57^{3} \cdot (1 - r)^{3}}{1 - r^{3}} = 9747 \frac{(1 - r)^{3}}{(1 - r) (1 + r^{2} + r)} = \frac{9747}{57^{3}} \frac{(1 - r)^{2}}{1 + r^{2} + r} = \frac{1}{19} \frac{1 + r^{2} - 2 r}{1 + r^{2} + r} = \frac{1}{19} 19 (1 + r^{2} - 2 r) = 1 + r^{2} + r 18 r^{2} - 39 r + 18 = 0 6 r^{2} - 13 r + 6 = 0 6 r^{2} - 9 r - 4 r + 6 = 0 (2 r - 3) (3 r - 2) = 0$

therefore r=3/2 or r=2/3 since r<1.

r=2/3

substitute in equation (1):

$\frac{a}{1 - \frac{2}{3}} = 57 3 a = 57 a = \frac{57}{3} = 19$

thus the first term of the GP is 19 and the common ratio is 2/3.

the GP is $19, \frac{38}{3}, \frac{76}{9}, a n d s o o n ...........$