If *S* denotes the sum of an infinite G.P. *S*_{1} denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively

\[\frac{2S S_1}{S^2 + S_1}\text { and } \frac{S^2 - S_1}{S^2 + S_1}\]

#### Solution

\[S = \frac{a}{\left( 1 - r \right)} . . . . . . . (i)\]

\[\text { And }, S_1 = \frac{a^2}{\left( 1 - r^2 \right)} \]

\[ \Rightarrow S_1 = \frac{a^2}{\left( 1 - r \right)\left( 1 + r \right)} . . . . . . . (ii)\]

\[\text { Now, putting the value of a in equation (ii) from equation } (i): \]

\[ S_1 = \frac{S^2 \left( 1 - r \right)^2}{\left( 1 - r \right)\left( 1 + r \right)}\]

\[ \Rightarrow S_1 = \frac{S^2 \left( 1 - r \right)}{\left( 1 + r \right)}\]

\[ \Rightarrow S_1 \left( 1 + r \right) = S^2 \left( 1 - r \right)\]

\[ \Rightarrow r\left( S_1 + S^2 \right) = S^2 - S_1 \]

\[ \Rightarrow r = \frac{\left( S^2 - S_1 \right)}{\left( S_1 + S^2 \right)}\]

\[\text { Putting the value of r in equation }(i): \]

\[ \Rightarrow a = S\left( 1 - r \right)\]

\[ \Rightarrow a = S\left( 1 - \frac{\left( S^2 - S_1 \right)}{\left( S_1 + S^2 \right)} \right)\]

\[ \Rightarrow a = S\left( \frac{\left( S_1 + S^2 \right) - \left( S^2 - S_1 \right)}{\left( S_1 + S^2 \right)} \right)\]

\[ \Rightarrow a = \frac{2 {SS}_1}{\left( S_1 + S^2 \right)}\]