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Geometric Progression

In a G.P., T ...

Question

In a G.P., $T_{2} + T_{5} = 216$ and $T_{4} : T_{6} = 1 : 4$ and all the terms are integers, then the first term is

14

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16

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Solution

The correct option is B $12$
$a r (1 + r^{3}) = 216$ and $\frac{a r^{3}}{a r^{5}} = \frac{1}{4}$
$\Rightarrow r^{2} = 4 \Rightarrow r = \pm 2$
when $r = 2$ then $2 a (9) = 216 \Rightarrow a = 12$
when $r = - 2$ , then $- 2 a (1 - 8) = 216$
$∴ a = \frac{216}{14} = \frac{108}{7}$ , which is not an integer.

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