Question

The sum of infinite terms of a geometric progression is 15 and the sum of their squares is 45. The value of common ratio is​

Answer 1

Let ,

• The first term and common ratio of GP be a and r

First Condition

The sum of infinite terms of a geometric progression is 15

We know that ,

The sum of infinite terms of GP is given by

$\boxed{ \sf{S_{n} = \frac{a}{(1 - r)} }}$

Thus ,

$\sf \mapsto 15 = \frac{a}{(1 - r)} \\ \\ \sf{ Squaring \: both \: sides \: , \: we \: get} \\ \\\sf \mapsto 225 = \frac{ {(a)}^{2} }{ {(1 - r)}^{2} } \: - - - \: (i)$

Second Condition

The sum of squares of infinite terms of given GP is 45

Thus ,

$\sf \mapsto 45 = \frac{ {(a)}^{2} }{1 - {(r)}^{2} } \\ \\\sf \mapsto 45 = \frac{ {(a)}^{2} }{(1 - r)(1 + r)} \: - - - \: (ii)$

Dividing eq (i) by eq (ii) , we get

$\sf \mapsto \frac{(1 + r)}{(1 - r)} = 5 \\ \\\sf \mapsto 1 + r = 5 - 5r \\ \\\sf \mapsto 6r = 4 \\ \\\sf \mapsto r = \frac{2}{3}$

$\sf \therefore \underline{The \: common \: ratio \: of \: given \: GP \: is \: \frac{2}{3} }$

Answer 2

Let ,

The first common ratio of GP will be a and r

First Condition

The sum of infinite terms in a geometric progression is 15

We know that ,

The sum of infinite terms of GP

$\boxed{ \sf{S_{n} = \frac{a}{(1 - r)} }}$

Thus,

$\sf \mapsto 15 = \frac{a}{(1 - r)}$

$\sf{ Squaring \: both \: sides \: , \: we \: get}$

$\sf \mapsto 225 = \frac{ {(a)}^{2} }{ {(1 - r)}^{2} } \: - - - \: (i)$

Second Condition

The sum of squares of infinite terms of given GP is 45

Thus ,

$\sf \mapsto 45 = \frac{ {(a)}^{2} }{1 - {(r)}^{2} }$

$\sf \mapsto 45 = \frac{ {(a)}^{2} }{(1 - r)(1 + r)} \: - - - \: (ii)$

Dividing eq (i) by eq (ii) , we get

$\sf \mapsto \frac{(1 + r)}{(1 - r)} = 5$

$\sf \mapsto 1 + r = 5 - 5r$

$\sf \mapsto 6r = 4$

$\sf \mapsto r = \frac{2}{3}$

$\sf \therefore \underline{The \: common \: ratio \: of \: given \: GP \: is \: \frac{2}{3} }$