If the sum of an infinite decreasing G.P. is 3 and the sum of the squares of its term is \[\frac{9}{2}\], then write its first term and common difference.

#### Solution

Let us take a G.P. whose first term is a and common difference is r.

\[\therefore S_\infty = \frac{a}{1 - r} \]

\[ \Rightarrow \frac{a}{1 - r} = 3 . . . . . . . \left( i \right)\]

\[\text { And, sum of the terms of the G . P } . a^2 , \left( ar \right)^2 , \left( a r^2 \right)^2 , . . . \infty : \]

\[S _\infty = \frac{a^2}{1 - r^2} \]

\[ \Rightarrow \frac{a^2}{1 - r^2} = \frac{9}{2} . . . . . . . \left( ii \right)\]

\[ \Rightarrow 2 a^2 = 9\left( 1 - r^2 \right) \]

\[ \Rightarrow 2 \left[ 3\left( 1 - r \right) \right]^2 = 9 - 9 r^2 \left[ \text { From } \left( i \right) \right]\]

\[ \Rightarrow 18\left( 1 + r^2 - 2r \right) = 9 - 9 r^2 \]

\[ \Rightarrow 18 - 9 + 18 r^2 + 9 r^2 - 36r = 0\]

\[ \Rightarrow 27 r^2 - 36r + 9 = 0\]

\[ \Rightarrow 3\left( 9 r^2 - 12r + 3 \right) = 0\]

\[ \Rightarrow 9 r^2 - 12r + 3 = 0\]

\[ \Rightarrow 9 r^2 - 9r - 3r + 3 = 0\]

\[ \Rightarrow 9r\left( r - 1 \right) - 3\left( r - 1 \right) = 0\]

\[ \Rightarrow \left( 9r - 3 \right)\left( r - 1 \right) = 0\]

\[ \Rightarrow r = \frac{1}{3} \text { and } r = 1 . \]

\[\text { But, r = 1 is not possible } . \]

\[ \therefore r = \frac{1}{3}\]

\[\text { Now, putting } r = \frac{1}{3} \text { in } \frac{a}{1 - r} = 3: \]

\[a = 3\left( 1 - \frac{1}{3} \right)\]

\[ \Rightarrow a = 3 \times \frac{2}{3} = 2\]