If the sum of the first $10$ terms of an arithmetic progression is $4$ times the sum of its first $5$ terms,then the ratio of its first term to the common difference is......

The sum of all two-digit positive numbers which,when divided by $7$,yield $2$ or $5$ as a remainder is:

If $a_r > 0, r \in N$ and $a_1, a_2, a_3, ..., a_{2n}$ are in an arithmetic progression,then $\frac{a_1 + a_{2n}}{\sqrt{a_1} + \sqrt{a_2}} + \frac{a_2 + a_{2n-1}}{\sqrt{a_2} + \sqrt{a_3}} + \frac{a_3 + a_{2n-2}}{\sqrt{a_3} + \sqrt{a_4}} + ... + \frac{a_n + a_{n+1}}{\sqrt{a_n} + \sqrt{a_{n+1}}} = ?$

If $\sum_{r=1}^n(2r+1)=440$,then $n = \ldots$.

The sums of $n$ terms of two arithmetic series are in the ratio $(2n + 3) : (6n + 5)$. Then the ratio of their $13^{th}$ terms is:

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