There are $15$ terms in an arithmetic progression. Its first term is $5$ and their sum is $390$. The middle term is

If the ratio of the sum of $n$ terms of two $A.P.'s$ be $(7n + 1):(4n + 27)$, then the ratio of their ${11^{th}}$ terms will be

The sum of integers from $1$ to $100$ that are divisible by $2$ or $5$ is

Let $V_{\mathrm{r}}$ denote the sum of the first $\mathrm{r}$ terms of an arithmetic progression $(A.P.)$ whose first term is $\mathrm{r}$ and the common difference is $(2 \mathrm{r}-1)$. Let

$T_{\mathrm{I}}=V_{\mathrm{r}+1}-V_{\mathrm{I}}-2 \text { and } \mathrm{Q}_{\mathrm{I}}=T_{\mathrm{r}+1}-\mathrm{T}_{\mathrm{r}} \text { for } \mathrm{r}=1,2, \ldots$

$1.$  The sum $V_1+V_2+\ldots+V_n$ is

$(A)$ $\frac{1}{12} n(n+1)\left(3 n^2-n+1\right)$

$(B)$ $\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$

$(C)$ $\frac{1}{2} n\left(2 n^2-n+1\right)$

$(D)$ $\frac{1}{3}\left(2 n^3-2 n+3\right)$

$2.$  $\mathrm{T}_{\mathrm{T}}$ is always

$(A)$ an odd number $(B)$ an even number

$(C)$ a prime number $(D)$ a composite number

$3.$  Which one of the following is a correct statement?

$(A)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $5$

$(B)$ $\mathrm{Q}_1, \mathrm{Q}_2, \mathrm{Q}_3, \ldots$ are in $A.P.$ with common difference $6$

$(C)$ $\mathrm{Q}_1, \mathrm{Q}_2, \mathrm{Q}_3, \ldots$ are in $A.P.$ with common difference $11$

$(D)$ $Q_1=Q_2=Q_3=\ldots$

Give the answer question $1,2$ and $3.$

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 :

If the sum of $n$ terms of an $A.P.$ is $\left(p n+q n^{2}\right),$ where $p$ and $q$ are constants, find the common difference.