Question

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

Answer 1

AnsWer:

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

$\qquad \sf \: Sum = 57 \implies \frac{a}{1 - r} = 57...(i) \\ \\ \qquad \sf \: Sum \: of \: the \: cubes = 9747 \\ \\ \implies\qquad \sf \: {a}^{3} - {a}^{3} {r}^{3} + {a}^{3} {r}^{6} + \dots = 9747 \\ \\ \implies \qquad \sf \: \frac{ {a}^{3} }{1 - {r}^{3} } = 9747...(ii)$

Dividing the cube of (i) by (ii), we get

$\qquad \sf \frac{ {a}^{3} }{ {(1 - r)}^{3} } \times \frac{ {(1 - r)}^{3} }{ {a}^{3} } = \frac{ {(57)}^{3} }{9747 } \\ \\ \implies \sf \frac{1 - {r}^{3} }{ {(1 - r)}^{3} } = 19 \\ \\ \implies \sf \frac{1 + r + {r}^{2} }{ {(1 - r)}^{2} } = 19 \\ \\ \implies \sf \: 18 {r}^{2} - 39r + 18 = 0 \\ \\ \implies \sf \: (3r - 2)(6r - 9) = 0 \\ \\ \implies \sf \: r = 2/3 \: \: or, \: r = 3/2 \\ \\ \implies \sf \: r = 2/3$

Putting r = 2/3 in (i), we get

$\qquad \tt \: \frac{a}{1 - (2/3)} = 57 \implies \: a = 19$

Hence, the G.P. is 19, 38/3, 76/9,...