If the first term of an $A.P.$ is $3$ and the sum of its first $25$ terms is equal to the sum of its next $15$ terms, then the common difference of this $A.P.$ is :

If the sum of the series $2 + 5 + 8 + 11............$ is $60100$, then the number of terms are

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$, the number of terms is

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 -

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 $a_1, a_2, a_3, \ldots, a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i, 1 \leq p \leq 100$. For any integer $n$ with $1 \leq n \leq 20$, let $m=5 n$. If $\frac{S_{m m}}{S_n}$ does not depend on $n$, then $a_2$ is