Suppose the sum of the first $m$ terms of an arithmetic progression is $n$ and the sum of its first $n$ terms is $m$, where $m \neq n$. Then, the sum of the first $(m+n)$ terms of the arithmetic progression is

Let $a_{1}, a_{2}, \ldots \ldots, a_{21}$ be an $A.P.$ such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}$. If the sum of this AP is $189,$ then  $a_{6} \mathrm{a}_{16}$ 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 $7^{\text {th }}$ and $(m-1)^{\text {th }}$ numbers is $5: 9 .$ Find the value of $m$

The ${n^{th}}$ term of an $A.P.$ is $3n - 1$.Choose from the following the sum of its first five terms

Consider a sequence whose sum of first $n$ -terms is given by $S_n = 4n^2 + 6n, n \in N$, then $T_{15}$ of this sequence is -

The number of terms common to the two A.P.'s $3,7,11, \ldots ., 407$ and $2,9,16, \ldots . .709$ is