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

$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped out on the second day,$4$ more workers dropped out on the third day,and so on. It takes $8$ more days to finish the work now. The number of days in which the work was completed is:

The product of three geometric means between $4$ and $\frac{1}{4}$ will be

If $a, b, c$ are in $G.P.$,then

If $a_1, a_2, a_3, ..., a_n$ are in $A.P.$,where $a_i > 0$ for all $i$,then the value of $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = $

The sum of $24$ terms of the following series $\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + \dots$ is