Maximum value of the sum of the arithmetic progression $50, 48, 46, 44, \dots$ is:

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

If $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ are $A.P.$ and $a_{1}=2, a_{10}=3, a_{1}b_{1}=1=a_{10}b_{10}$,then $a_{4}b_{4}$ is equal to

Between $1$ and $31$,$m$ numbers have been inserted in such a way that the resulting sequence is an $A.P.$ and the ratio of the $7^{\text{th}}$ and $(m-1)^{\text{th}}$ inserted numbers is $5:9$. Find the value of $m$.

Let $f: R \rightarrow R$ be such that for all $x \in R$,the terms $(2^{1+x}+2^{1-x})$,$f(x)$,and $(3^x+3^{-x})$ are in $A.P.$. Then the minimum value of $f(x)$ is:

In an $A.P.$,the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that the $20^{th}$ term is $-112$.