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

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Solution

The correct option is C
12

ar(1 + $r^{3}$ ) = 216 and $\frac{a r^{3}}{a r^{5}}$ = $\frac{1}{4}$

⇒ $r^{2}$ = 4 ⇒ r = 2, -2

When r = 2 then 2a(9) = 216 ∴ a = 12

When r = -2, then -2a(1 - 8) = 216

∴ a = $\frac{216}{14}$ = $\frac{108}{7}$ , which is not an integer.

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

T_{2} + T_{5} = 216

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T_{2} + T_{5} = 216

T_{4} : T_{6} = 1 : 4

T_{2} + T_{5} = 216

T_{4} : T_{6} = 1 : 4

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