Given sum of the first $n$ terms of an $A.P.$ is $2n + 3n^2.$ Another $A.P.$ is formed with the same first term and double of the common difference, the sum of $n$ terms of the new $A.P.$ is

If $x,y,z$ are in $A.P. $ and ${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$ and ${\tan ^{ - 1}}z$ are also in $A.P.$, then

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.

Five numbers are in $A.P.$, whose sum is $25$ and product is $2520 .$ If one of these five numbers is $-\frac{1}{2},$ then the greatest number amongst them is

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 $..........$.

Find the sum of all two digit numbers which when divided by $4,$ yields $1$ as remainder.