The first term of an $A.P.$ is $2$ and the common difference is $4$. The sum of its $40$ terms will be:

Let $a_1, a_2, a_3, \dots$ be an $A.P.$ such that $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^3}{q^3}$ where $p \neq q$. Then $\frac{a_6}{a_{21}}$ is equal to:

If the sum of the first $21$ terms of the series $\log _{9^{1 / 2}} x + \log _{9^{1 / 3}} x + \log _{9^{1 / 4}} x + \dots$ where $x > 0$ is $504$,then $x$ is equal to:

What is the sum of all two-digit numbers which give a remainder of $4$ when divided by $6$?

In an $A.P.$,if the $p^{\text{th}}$ term is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$,prove that the sum of the first $pq$ terms is $\frac{1}{2}(pq+1)$,where $p \neq q$.

The sum of the series $1 + 3 + 5 + 7 + \dots$ up to $n$ terms is equal to: