The solution of the equation $1 + a + a^2 + a^3 + \dots + a^x = (1 + a)(1 + a^2)(1 + a^4)$ is given by $x$ is equal to

How many terms of the $A.P.$ $1, 4, 7, \ldots$ are needed to give the sum $715$?

The first term of an $A.P.$ of consecutive integers is $p^2 + 1$. The sum of $(2p + 1)$ terms of this series can be expressed as:

If $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n}$ is the harmonic mean between $a$ and $b$,then the value of $n$ is

Sum the series $51 + 50 + 49 + \ldots + 21$.

If the sum of the first $15$ terms of the series ${\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + \dots$ is equal to $225k$,then $k$ is equal to