If $T_{1}, T_{2}, T_{8}$ of a certain $G . P$ be $x^{- 4}, x^{k}$ and $x^{52}$ respectively, then prove that $k = 4$ .

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Solution

If $r$ be the common ration, then $r = \frac{T_{2}}{T_{1}} = x^{k + 4}$
$T_{8} = {a r}^{7} = x^{- 4} x^{7} = x^{7 (k + 4)} = x^{52}$
$∴ 7 k + 28 - 4 = 52$
$∴ 7 k = 28$ or $k = 4$

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