The $n^{th}$ term of the series $\frac{1}{1} + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$ will be

The sum $1(1!) + 2(2!) + 3(3!) + ....+n (n!)$ equals

The common difference of the $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $2$ more than the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}.$ If $a_{40} = -159,$ $a_{100} = -399$ and $b_{100} = a_{70},$ then $b_{1}$ is equal to

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

The sum of an infinite geometric series is $3$. $A$ series,which is formed by squares of its terms,has the sum also $3$. The first series is

Given a sequence of $4$ numbers,the first three of which are in $G.P.$ and the last three are in $A.P.$ with a common difference of $6$. If the first and last terms in this sequence are equal,then the last term is: