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Q. Let $\{T_n\}$ be a sequence of integers in G.P. in which $T_4:T_6=1:4$ and $T_2+T_5=216$ . then $T_1$ is

Sequences and Series

Solution:

$\frac{T_{4}}{T_{6}} = \frac{aR^{3}}{aR^{5}} = \frac{1}{4} $
$ \Rightarrow R^{2} = 4$
$ \Rightarrow R=2 $
$T_{2} +T_{5} = 216 $
$ \Rightarrow aR +aR^{4} = 216 $
$\Rightarrow aR \left(1+R^{3}\right) = 216$
$ \Rightarrow a\cdot2\left(1+8\right) = 216$
$ \Rightarrow a\left(1\right) = \frac{108}{9} = 12$.
$Thus T_{1} = a=12.$