Question

The sum of inifinte g.p is. 15and the sum of their square is 45.then find the series

Answer 1

Let the 1st term of GP be a and common ratio be r
Given,
$\frac{a}{1 - r} = 15 - - - (1) \\ and \: \: \frac{ {a}^{2} }{1 - {r}^{2} } = 45 - - - (2)\\ squaring \: (1) \: \: {( \frac{a}{1 - r})}^{2} = 225 \\ \frac{ {a}^{2} }{ {(1 - r)}^{2} } = 225 - - - (3) \\ \\ \frac{(3)}{(2)} \: gives \: \: \frac{1 - {r}^{2} }{ {(1 - r)}^{2} } = \frac{225}{45} = 5 \\ \frac{1 + r}{1 - r} = 5 \\ 1 + r = 5 - 5r \\ 6r = 4 \\ r = \frac{2}{3} \\ \\ from \: (1) \: \: a = 15 \times (1 - r) \\ = 15 \times (1 - \frac{2}{3} ) = 5 \\ \\ thus \: \: series \: \: is \\ 5 + \frac{10}{3} + \frac{20}{9} + .....$