Step-by-Step Solution

1. Understand the Definitions

We are given several definitions based on a variable r:

• $V_r$: The sum of the first r terms of an Arithmetic Progression (A.P.).
• For this A.P., the first term is a = r and the common difference is d = 2r - 1.
• $T_r = V_{r+1} - V_r - 2$.
• $Q_r = T_{r+1} - T_r$.

Our goal is to determine the nature of the sequence $Q_1, Q_2, Q_3, ...$.

2. Find an Expression for $V_r$

The sum of the first n terms of an A.P. is given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$. For $V_r$, we have n=r, a=r, and d=2r-1. Substituting these into the formula: $V_r = \frac{r}{2}[2(r) + (r-1)(2r-1)]$ $V_r = \frac{r}{2}[2r + (2r^2 - 2r - r + 1)]$ $V_r = \frac{r}{2}[2r + 2r^2 - 3r + 1]$ $V_r = \frac{r}{2}[2r^2 - r + 1]$

3. Find an Expression for $T_r$

$T_r$ is defined as $T_r = V_{r+1} - V_r - 2$. First, let's find the expression for $V_{r+1}$ by substituting r+1 for r in the expression for $V_r$: $V_{r+1} = \frac{r+1}{2}[2(r+1)^2 - (r+1) + 1]$ $V_{r+1} = \frac{r+1}{2}[2(r^2 + 2r + 1) - r - 1 + 1]$ $V_{r+1} = \frac{r+1}{2}[2r^2 + 4r + 2 - r]$ $V_{r+1} = \frac{r+1}{2}[2r^2 + 3r + 2]$

Now, let's compute the difference $V_{r+1} - V_r$: $V_{r+1} - V_r = \frac{1}{2}[(r+1)(2r^2 + 3r + 2) - r(2r^2 - r + 1)]$ $V_{r+1} - V_r = \frac{1}{2}[(2r^3 + 3r^2 + 2r + 2r^2 + 3r + 2) - (2r^3 - r^2 + r)]$ $V_{r+1} - V_r = \frac{1}{2}[2r^3 + 5r^2 + 5r + 2 - 2r^3 + r^2 - r]$ $V_{r+1} - V_r = \frac{1}{2}[6r^2 + 4r + 2]$ $V_{r+1} - V_r = 3r^2 + 2r + 1$

Now, substitute this result into the expression for $T_r$: $T_r = (V_{r+1} - V_r) - 2$ $T_r = (3r^2 + 2r + 1) - 2$ $T_r = 3r^2 + 2r - 1$

4. Find an Expression for $Q_r$

$Q_r$ is defined as $Q_r = T_{r+1} - T_r$. First, let's find the expression for $T_{r+1}$ by substituting r+1 for r in the expression for $T_r$: $T_{r+1} = 3(r+1)^2 + 2(r+1) - 1$ $T_{r+1} = 3(r^2 + 2r + 1) + 2r + 2 - 1$ $T_{r+1} = 3r^2 + 6r + 3 + 2r + 1$ $T_{r+1} = 3r^2 + 8r + 4$

Now, compute $Q_r$: $Q_r = T_{r+1} - T_r$ $Q_r = (3r^2 + 8r + 4) - (3r^2 + 2r - 1)$ $Q_r = 3r^2 + 8r + 4 - 3r^2 - 2r + 1$ $Q_r = 6r + 5$

5. Analyze the Sequence $Q_r$

The general term of the sequence is $Q_r = 6r + 5$. To determine if this is an A.P., we check the difference between consecutive terms: $Q_{r+1} - Q_r = [6(r+1) + 5] - [6r + 5]$ $= (6r + 6 + 5) - (6r + 5)$ $= 6r + 11 - 6r - 5$ $= 6$ Since the difference between consecutive terms is a constant value of 6, the sequence $Q_1, Q_2, Q_3, ...$ is an A.P. with a common difference of 6.

Let's find the first few terms to confirm:

• $Q_1 = 6(1) + 5 = 11$
• $Q_2 = 6(2) + 5 = 17$
• $Q_3 = 6(3) + 5 = 23$ The sequence is 11, 17, 23, ..., which is indeed an A.P. with a common difference of 6.

6. Evaluate the Options

• A: ... common difference 5 (Incorrect)
• B: ... common difference 6 (Correct)
• C: ... common difference 11 (Incorrect, 11 is the first term)
• D: $Q_1 = Q_2 = Q_3, ...$ (Incorrect)

The correct statement is that $Q_1, Q_2, Q_3, ...$ are in A.P. with a common difference of 6.