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

Consider four consecutive integers in an increasing arithmetic progression. One of these integers is equal to the sum of the squares of the other three. What is the sum of all the numbers?

The common difference of the $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $2$ more than the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$. If $a_{40} = -159$,$a_{100} = -399$ and $b_{100} = a_{70}$,then $b_{1}$ is equal to:

The maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + \dots$ 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 :

Suppose that the number of terms in an $A.P.$ is $2k$,where $k \in N$. If the sum of all odd-positioned terms of the $A.P.$ is $40$,the sum of all even-positioned terms is $55$,and the last term of the $A.P.$ exceeds the first term by $27$,then $k$ is equal to: