If the sum of an infinite GP a, ar, ar^{2}, a | vedclass

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(B) The sum of the infinite $GP$ is given by $\frac{a}{1-r} = 15 \dots (i)$.The series formed by the squares of the terms is $a^{2}, a^{2}r^{2}, a^{2}

Vedclass · Accepted Answer

$\frac{1}{2}$

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(B) The sum of the infinite $GP$ is given by $\frac{a}{1-r} = 15 \dots (i)$.The series formed by the squares of the terms is $a^{2}, a^{2}r^{2}, a^{2}r^{4}, \dots$,which is also a $GP$ with first term $a^{2}$ and common ratio $r^{2}$.The sum of this series is $\frac{a^{2}}{1-r^{2}} = 150$.We can rewrite this as $\frac{a}{1-r} \cdot \frac{a}{1+r} = 150$.Substituting $(i)$ into this equation,we get $15 \cdot \frac{a}{1+r} = 150$,which implies $\frac{a}{1+r} = 10 \dots (ii)$.Dividing $(i)$ by $(ii)$,we get $\frac{1+r}{1-r} = \frac{15}{10} = \frac{3}{2}$.Solving for $r$: $2 + 2r = 3 - 3r$ $\Rightarrow 5r = 1$ $\Rightarrow r = \frac{1}{5}$.Substituting $r = \frac{1}{5}$ into $(i)$: $\frac{a}{1 - 1/5} = 15$ $\Rightarrow \frac{a}{4/5} = 15$ $\Rightarrow a = 15 \cdot \frac{4}{5} = 12$.The series $ar^{2}, ar^{4}, ar^{6}, \dots$ is a $GP$ with first term $A = ar^{2}$ and common ratio $R = r^{2}$.Sum $= \frac{ar^{2}}{1-r^{2}} = \frac{12 \cdot (1/5)^{2}}{1 - (1/5)^{2}} = \frac{12 \cdot (1/25)}{1 - 1/25} = \frac{12/25}{24/25} = \frac{12}{24} = \frac{1}{2}$.

If the sum of an infinite $GP$ $a, ar, ar^{2}, ar^{3}, \ldots$ is $15$ and the sum of the squares of its each term is $150$,then the sum of $ar^{2}, ar^{4}, ar^{6}, \ldots$ is:

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