Arithmetic Progression Common Difference of Sequence Qn

To determine the correct statement regarding the sequence $Q_1, Q_2, Q_3, \dots$, we recall the definitions from the passage:

Let $V_r$ denote the sum of the first $r$ terms of an arithmetic progression whose first term is $r$ and common difference is $2r - 1$.

The sum of the first $r$ terms of this progression is given by:
$V_r = \frac{r}{2} \left[2(r) + (r - 1)(2r - 1)\right]$
$V_r = \frac{r}{2} \left[2r + 2r^2 - 3r + 1\right] = \frac{r}{2} \left(2r^2 - r + 1\right) = r^3 - \frac{1}{2}r^2 + \frac{1}{2}r$

Replacing $r$ with $r+1$, we have:
$V_{r+1} = \frac{r+1}{2} \left[2(r+1)^2 - (r+1) + 1\right] = \frac{r+1}{2} \left(2r^2 + 3r + 2\right)$
$V_{r+1} = (r+1)\left(r^2 + \frac{3}{2}r + 1\right) = r^3 + \frac{5}{2}r^2 + \frac{5}{2}r + 1$

Now, $T_r$ is defined as:
$T_r = V_{r+1} - V_r - 2$
$T_r = \left(r^3 + \frac{5}{2}r^2 + \frac{5}{2}r + 1\right) - \left(r^3 - \frac{1}{2}r^2 + \frac{1}{2}r\right) - 2$
$T_r = 3r^2 + 2r + 1 - 2 = 3r^2 + 2r - 1$

The term $Q_r$ is defined as:
$Q_r = T_{r+1} - T_r$

Evaluating $T_{r+1}$:
$T_{r+1} = 3(r+1)^2 + 2(r+1) - 1 = 3(r^2 + 2r + 1) + 2r + 2 - 1 = 3r^2 + 8r + 4$

Thus:
$Q_r = (3r^2 + 8r + 4) - (3r^2 + 2r - 1) = 6r + 5$

For successive terms of the sequence $\{Q_r\}$:
$Q_{r+1} - Q_r = [6(r+1) + 5] - (6r + 5) = 6$

Since the difference between any two consecutive terms $Q_{r+1} - Q_r$ is a constant equal to $6$, the sequence $Q_1, Q_2, Q_3, \dots$ forms an Arithmetic Progression (A.P.) with a common difference of $6$.

Therefore, the correct statement is that $Q_1, Q_2, Q_3, \dots$ are in A.P. with common difference 6.