Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms.

Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1{ }^2-\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2-\mathrm{a}_4{ }^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}{ }^2-\mathrm{a}_{2 \mathrm{k}}{ }^2$.

If $\mathrm{A}_3=-153, \mathrm{~A}_5=-435$ and $\mathrm{a}_1{ }^2+\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2=66$, then $\mathrm{a}_{17}-\mathrm{A}_7$ is equal to....................

The number of terms common between the two series $2 + 5 + 8 +.....$ upto $50$ terms and the series $3 + 5 + 7 + 9.....$ upto $60$ terms, is

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 ${m^{th}}$ terms of the series $63 + 65 + 67 + 69 + .........$ and $3 + 10 + 17 + 24 + ......$ be equal, then $m = $

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

The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + ........$ to $9$ terms is