Let V_r denote the sum of the first r terms o | vedclass

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(B) $1.$ Given $V_r$ is the sum of $r$ terms of an $A.P.$ with first term $a=r$ and common difference $d=2r-1$.$V_r = \frac{r}{2}[2a + (r-1)d] = \frac

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$B, D, A$

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(B) $1.$ Given $V_r$ is the sum of $r$ terms of an $A.P.$ with first term $a=r$ and common difference $d=2r-1$.$V_r = \frac{r}{2}[2a + (r-1)d] = \frac{r}{2}[2r + (r-1)(2r-1)] = \frac{r}{2}[2r + 2r^2 - 3r + 1] = \frac{r}{2}[2r^2 - r + 1] = r^3 - \frac{1}{2}r^2 + \frac{1}{2}r$.Sum $\sum_{r=1}^n V_r = \sum r^3 - \frac{1}{2} \sum r^2 + \frac{1}{2} \sum r = [\frac{n(n+1)}{2}]^2 - \frac{1}{2} \frac{n(n+1)(2n+1)}{6} + \frac{1}{2} \frac{n(n+1)}{2}$.$= \frac{n(n+1)}{12} [3n(n+1) - (2n+1) + 3] = \frac{n(n+1)}{12} [3n^2 + 3n - 2n - 1 + 3] = \frac{n(n+1)}{12} (3n^2 + n + 2)$.$2.$ $T_r = V_{r+1} - V_r - 2$. Since $V_{r+1} - V_r$ is the $(r+1)$-th term of the $A.P.$,$V_{r+1} - V_r = a + ((r+1)-1)d = r + r(2r-1) = r + 2r^2 - r = 2r^2$.Thus,$T_r = 2r^2 - 2 = 2(r^2 - 1) = 2(r-1)(r+1)$.For $r=1, T_1 = 0$. For $r > 1$,$T_r$ is a product of at least three factors (including $2$),so it is a composite number.$3.$ $Q_r = T_{r+1} - T_r = [2(r+1)^2 - 2] - [2r^2 - 2] = 2(r^2 + 2r + 1 - 1 - r^2 + 1) = 2(2r + 1) = 4r + 2$.$Q_1 = 6, Q_2 = 10, Q_3 = 14$. This is an $A.P.$ with common difference $4$. Wait,re-evaluating $V_{r+1}-V_r$: $V_{r+1}-V_r$ is the $(r+1)$-th term of the $A.P.$ with first term $r$ and $d=2r-1$. The $(r+1)$-th term is $r + r(2r-1) = 2r^2$. Correct. $T_r = 2r^2-2$. $Q_r = T_{r+1}-T_r = 2(r+1)^2 - 2r^2 = 2(2r+1) = 4r+2$. The common difference is $4$. Given the options,let's re-check the $V_r$ definition. If $V_r$ is sum of $r$ terms of $A.P.$ with first term $r$ and $d=2r-1$,then $V_r = r^3 - 0.5r^2 + 0.5r$. The calculation leads to $B, D, B$ based on standard interpretation.

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