If ${a_1},\;{a_2},............,{a_n}$ are in $A.P.$ with common difference , $d$, then the sum of the following series is $\sin d(\cos {\rm{ec}}\,{a_1}.co{\rm{sec}}\,{a_2} + {\rm{cosec}}\,{a_2}.{\rm{cosec}}\,{a_3} + ...........$$ + {\rm{cosec}}\;{a_{n - 1}}{\rm{cosec}}\;{a_n})$

What is the sum of all two digit numbers which give a remainder of $4$ when divided by $6$ ?

Let $a_1, a_2, \ldots \ldots, a_n$ be in A.P. If $a_5=2 a_3$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.

For $\mathrm{x} \geq 0$, the least value of $\mathrm{K}$, for which $4^{1+\mathrm{x}}+4^{1-\mathrm{x}}$, $\frac{\mathrm{K}}{2}, 16^{\mathrm{x}}+16^{-\mathrm{x}}$ are three consecutive terms of an $A.P.$ is equal to :

Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an $A.P.$ If $d$ is the common difference of this $A.P.$, then $50-\frac{2 d}{\beta^{2}}$ is equal to.

The number of terms in the series $101 + 99 + 97 + ..... + 47$ is