If $2^{10}+2^{9} \cdot 3^{1}+28 \cdot 3^{2}+\ldots+2 \cdot 3^{9}+3^{10}=S -211$ then $S$ is equal to
$\frac{3^{11}}{2}+2^{10}$
$3^{11}-2^{12}$
$3^{11}$
$2 \cdot 3^{11}$
If $x, {G_1},{G_2},\;y$ be the consecutive terms of a $G.P.$, then the value of ${G_1}\,{G_2}$ will be
Find the $10^{\text {th }}$ and $n^{\text {th }}$ terms of the $G.P.$ $5,25,125, \ldots$
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
Let $n \geq 3$ and let $C_1, C_2, \ldots, C_n$, be circles with radii $r_1, r_2, \ldots, r_n$, respectively. Assume that $C_i$ and $C_{i+1}$ touch externally for $1 \leq i \leq n-1$. It is also given that the $X$-axis and the line $y=2 \sqrt{2} x+10$ are tangential to each of the circles. Then, $r_1, r_2, \ldots, r_n$ are in
If the sum of an infinite $GP$ $a, ar, ar^{2}, a r^{3}, \ldots$ is $15$ and the sum of the squares of its each term is $150 ,$ then the sum of $\mathrm{ar}^{2}, \mathrm{ar}^{4}, \mathrm{ar}^{6}, \ldots$ is :