If the sum of $n$ terms of an $A.P$. is $2{n^2} + 5n$, then the ${n^{th}}$ term will be

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 $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an $A.P.$ are $a, b, c,$ respectively. Show that $(q-r) a+(r-p) b+(p-q) c=0$

Given sum of the first $n$ terms of an $A.P.$ is $2n + 3n^2.$ Another $A.P.$ is formed with the same first term and double of the common difference, the sum of $n$ terms of the new $A.P.$ is

If ${m^{th}}$ terms of the series $63 + 65 + 67 + 69 + .........$ and $3 + 10 + 17 + 24 + ......$ be equal, then $m = $

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