Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in N$, be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+k, \in N$. If $c_2=5$ and $c_3=13 / 4$ then $\sum \limits_{k=1}^{\infty} c_k - \left(12 a _6+8 b _4\right)$ is equal to
$9$
$18$
$20$
$22$
The sum of infinity of a geometric progression is $\frac{4}{3}$ and the first term is $\frac{3}{4}$. The common ratio is
If $G$ be the geometric mean of $x$ and $y$, then $\frac{1}{{{G^2} - {x^2}}} + \frac{1}{{{G^2} - {y^2}}} = $
The sum of infinite terms of a $G.P.$ is $x$ and on squaring the each term of it, the sum will be $y$, then the common ratio of this series is
If $n$ geometric means be inserted between $a$ and $b$ then the ${n^{th}}$ geometric mean will be
$0.5737373...... = $