Find the sum to $n$ terms of the $A.P.,$ whose $k^{\text {th }}$ term is $5 k+1$

Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots$ be an $A.P.$ If $\frac{a_{1}+a_{2}+\ldots+a_{10}}{a_{1}+a_{2}+\ldots+a_{p}}=\frac{100}{p^{2}}, p \neq 10$, then $\frac{a_{11}}{a_{10}}$ is equal to :

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

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}} }} = $

If the sum of $n$ terms of an $A.P$. is $2{n^2} + 5n$, then the ${n^{th}}$ term will be

The sum of all those terms, of the anithmetic progression $3,8,13, \ldots \ldots .373$, which are not divisible by $3$,is equal to $.......$.