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:

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 12
We know that $n t h$ term of an GP is given by $a r^{n - 1}$
Since, $a r + a r^{4} = 216$ and $\frac{a r^{3}}{a r^{5}} = \frac{1}{4}$
$\Rightarrow r^{2} = 4 \Rightarrow r = 2, - 2$
When $r = 2,$ then $2 a . (9) = 216$
$∴ a = 12$
When $r = - 2,$ then $- 2 a (1 - 8) = 216$
Thus, $a = \frac{216}{14} = \frac{108}{7}$ , which is not an integer.
Therefore, $a = 12$

Suggest Corrections

Similar questions

T_{2} + T_{5} = 216

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

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

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

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

t_{0}, t_{1}, t_{2}, t_{n}

(x + a i)^{n}

(t_{0} - t_{2} + t_{4} - t_{6} + \dots .)^{2} + (t_{1} - t_{3} + t_{5} \dots .)^{2} =

Join BYJU'S Learning Program