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(A) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 3$ .....$(i)$The squares of

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$3/2, 1/2$

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(A) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 3$ .....$(i)$The squares of the terms form a new $G.P.$ with first term $a^2$ and common ratio $r^2$. The sum of this infinite $G.P.$ is $\frac{a^2}{1-r^2} = 3$ .....(ii)From (ii),we have $\frac{a^2}{(1-r)(1+r)} = 3$. Substituting $(i)$ into this,we get $\frac{a}{1+r} = 1$,which implies $a = 1+r$.Substituting $a = 1+r$ into $(i)$: $\frac{1+r}{1-r} = 3$.$1+r = 3 - 3r$ $\Rightarrow 4r = 2$ $\Rightarrow r = 1/2$.Using $a = 1+r$,we get $a = 1 + 1/2 = 3/2$.Thus,the first term is $3/2$ and the common ratio is $1/2$.

If the sum of infinite terms of a $G.P.$ is $3$ and the sum of the squares of its terms is $3$,then its first term and common ratio are:

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