Let ${a_1},{a_2},\;.\;.\;.\;.,{a_{49}}$ be in $A.P.$ such that $\mathop \sum \limits_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m,$ then $m = \;\;..\;.\;.\;.\;$

Let $a_1, a_2, \ldots \ldots, a_n$ be in A.P. If $a_5=2 a_3$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.

Let $a_1 , a_2, a_3, .... , a_n$, be in $A.P$. If $a_3 + a_7 + a_{11} + a_{15} = 72$ , then the sum of its first $17$ terms is equal to

If sum of $n$ terms of an $A.P.$ is $3{n^2} + 5n$ and ${T_m} = 164$ then $m = $

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 $3.$

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