Consider two G.Ps. $2,2^{2}, 2^{3}, \ldots$ and $4,4^{2}, 4^{3}, \ldots$ of $60$ and $n$ terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\frac{225}{8}}$, then $\sum_{ k =1}^{ n } k (n- k )$ is equal to.
$560$
$1540$
$1330$
$2600$
If in an infinite $G.P.$ first term is equal to the twice of the sum of the remaining terms, then its common ratio is
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
Show that the products of the corresponding terms of the sequences $a,$ $ar,$ $a r^{2},$ $......a r^{n-1}$ and $A, A R, A R^{2}, \ldots, A R^{n-1}$ form a $G .P.,$ and find the common ratio.
There are two such pairs of non-zero real valuesof $a$ and $b$ i.e. $(a_1,b_1)$ and $(a_2,b_2)$ for which $2a+b,a-b,a+3b$ are three consecutive terms of a $G.P.$, then the value of $2(a_1b_2 + a_2b_1) + 9a_1a_2$ is-
Let $a, a r, a r^2, \ldots . . .$. be an infinite $G.P.$ If $\sum_{n=0}^{\infty} a^n=57$ and $\sum_{n=0}^{\infty} a^3 r^{3 n}=9747$, then $a+18 r$ is equal to :