Find the sum of $1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots$ to infinite terms.

If the sum of the $10$ terms of an $A.P.$ is $4$ times the sum of its $5$ terms,then the ratio of the first term to the common difference is:

The sum,$\sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4}$ is equal to

The $n^{th}$ term of an $A.P.$ is $3n - 1$. Choose from the following the sum of its first five terms.

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$,such that $B_n > A_n$ for all $n \geq p$,is

If $n$ geometric means are inserted between $a$ and $b$,then the $n^{th}$ geometric mean will be: