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# If the Sum of an Infinite Decreasing G.P. is 3 and the Sum of the Squares of Its Term is 9 2 , Then Write Its First Term and Common Difference. - Mathematics

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.

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#### Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow 9 r^2 - 9r - 3r + 3 = 0$

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

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

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

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

$\therefore r = \frac{1}{3}$

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

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

$\Rightarrow a = 3 \times \frac{2}{3} = 2$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Q 4 | Page 56
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