If the first term of a $G.P.$ is $5$ and the common ratio is $-5$,then which term is $3125$ (in $^{th}$)?

If $^nC_4, ^nC_5,$ and $^nC_6$ are in $A.P.,$ then $n$ can be:

The sum of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$ up to infinity is:

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

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

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,the number of terms is