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A $G.P.$ consists of an even number of terms. If the sum of all the terms is $5$ times the sum of terms occupying odd places, then find its common ratio.
$4$
$4$
$4$
$4$
Solution
Let the $G.P.$ be $T_{1}, T_{2}, T_{3}, T_{4} \ldots . T_{2 n}$
Number of terms $=2 n$
According to the given condition,
$T_{1}+T_{2}+T_{3}+\ldots .+T_{2 n}=5\left[T_{1}+T_{3}+\ldots .+T_{2 n-1}\right]$
$\Rightarrow T_{1}+T_{2}+T_{3}+\ldots .+T_{2 n}-5\left[T_{1}+T_{3}+\ldots . .+T_{2 n-1}\right]=0$
$\Rightarrow T_{2}+T_{4}+\ldots .+T_{2 n}=4\left[T_{1}+T_{3}+\ldots . .+T_{2 n-1}\right]$
Let the $G.P.$ be $a, a r, a r^{2}, a r^{3} \dots$
$\therefore \frac{\operatorname{ar}\left(r^{n}-1\right)}{r-1}=\frac{4 \times a\left(r^{n}-1\right)}{r-1}$
$\Rightarrow a r=4 a$
$\Rightarrow r=4$
Thus, the common ratio of the $G.P.$ is $4$