Question

If the n terms a₁,a₂........a_n are in A.P. with increment r, then the difference between the mean of their squares and square of their mean is 

• A. \(\frac{r^2[(n-1)^2-1]}{12}\)
• B. \(\frac{r^2}{12}\)
• C. \(\frac{r^2(n^2-1)}{12}\)
• D. \(\frac{n^2-1}{12}\)

Answer 1

The Correct Option is C

In an arithmetic progression (A.P.) with common difference $r$, the terms can be written as:

$a_1, a_1 + r, a_1 + 2r, \dots, a_1 + (n - 1)r$

The mean of these terms is:

$\text{Mean} = \dfrac{a_1 + a_1 + r + \dots + a_1 + (n - 1)r}{n}$ 

The mean of their squares is:

$\text{Mean of Squares} = \dfrac{(a_1)^2 + (a_1 + r)^2 + \dots + (a_1 + (n - 1)r)^2}{n}$

Now, the difference between the mean of squares and the square of the mean is:

$\text{Difference} = \dfrac{1}{n} \sum_{k=0}^{n-1} (a_1 + kr)^2 - \left(\dfrac{1}{n} \sum_{k=0}^{n-1} (a_1 + kr)\right)^2$

This expression represents the variance of the A.P. terms. Using known formulas:

• $\sum_{k=0}^{n-1} (a_1 + kr)^2 = n a_1^2 + a_1 r (n(n - 1)) + r^2 \dfrac{(n - 1)n(2n - 1)}{6}$
• $\left(\dfrac{1}{n} \sum_{k=0}^{n-1} (a_1 + kr)\right)^2 = \left(a_1 + \dfrac{(n - 1)r}{2}\right)^2$

After simplifying, we get:

Difference = $\dfrac{r^2(n^2 - 1)}{12}$

So, the correct answer is Option (C): $\boxed{\dfrac{r^2(n^2 - 1)}{12}}$

Answer 2

Step 1: Define the Terms of the A.P.

The terms of the A.P. can be written as:

\(a_i = a_1 + (i - 1)r\) for \(i = 1, 2, \dots, n\)

Step 2: Calculate the Mean of the Terms

The mean of the terms is:

\(\bar{a} = \frac{1}{n} \sum_{i=1}^n a_i = \frac{1}{n} \sum_{i=1}^n [a_1 + (i - 1)r]\)

\(\bar{a} = \frac{1}{n} \left[ \sum_{i=1}^n a_1 + \sum_{i=1}^n (i - 1)r \right] = \frac{1}{n} \left[ na_1 + r \sum_{i=1}^n (i - 1) \right]\)

We know that \(\sum_{i=1}^n (i - 1) = \sum_{i=0}^{n-1} i = \frac{(n - 1)n}{2}\), so

\(\bar{a} = \frac{1}{n} \left[ na_1 + r \frac{n(n - 1)}{2} \right] = a_1 + \frac{(n - 1)r}{2}\)

Step 3: Calculate the Mean of the Squares of the Terms

The mean of the squares of the terms is:

\(\frac{1}{n} \sum_{i=1}^n a_i^2 = \frac{1}{n} \sum_{i=1}^n [a_1 + (i - 1)r]^2\)

\(\frac{1}{n} \sum_{i=1}^n [a_1^2 + 2a_1(i-1)r + (i-1)^2 r^2] = \frac{1}{n} \left[ \sum_{i=1}^n a_1^2 + 2a_1r \sum_{i=1}^n (i - 1) + r^2 \sum_{i=1}^n (i - 1)^2 \right]\)

We know that \(\sum_{i=1}^n a_1^2 = na_1^2\), \(\sum_{i=1}^n (i - 1) = \frac{n(n - 1)}{2}\) and \(\sum_{i=1}^n (i - 1)^2 = \frac{(n-1)n(2n-1)}{6}\), so

\(\frac{1}{n} \left[ na_1^2 + 2a_1r \frac{n(n - 1)}{2} + r^2 \frac{(n - 1)n(2n - 1)}{6} \right] = a_1^2 + a_1r(n - 1) + \frac{r^2 (n - 1)(2n - 1)}{6}\)

Step 4: Calculate the Square of the Mean

The square of the mean is:

\(\left( \frac{1}{n} \sum_{i=1}^n a_i \right)^2 = \bar{a}^2 = \left( a_1 + \frac{(n - 1)r}{2} \right)^2 = a_1^2 + 2 a_1 \frac{(n - 1)r}{2} + \left(\frac{(n - 1)r}{2}\right)^2\)

\(\bar{a}^2 = a_1^2 + a_1 (n - 1)r + \frac{(n - 1)^2 r^2}{4}\)

Step 5: Find the Difference

The difference between the mean of the squares and the square of the mean is:

\(D = \frac{1}{n} \sum_{i=1}^n a_i^2 - \bar{a}^2 = \left[a_1^2 + a_1 r(n - 1) + \frac{r^2 (n - 1)(2n - 1)}{6} \right] - \left[a_1^2 + a_1 r(n - 1) + \frac{r^2 (n - 1)^2}{4}\right]\)

\(D = \frac{r^2 (n - 1)(2n - 1)}{6} - \frac{r^2 (n - 1)^2}{4}\)

\(D = r^2(n - 1) \left( \frac{2n - 1}{6} - \frac{n - 1}{4} \right) = r^2(n - 1) \left( \frac{2(2n - 1) - 3(n - 1)}{12} \right)\)

\(D = r^2 (n - 1) \left( \frac{4n - 2 - 3n + 3}{12} \right) = r^2 (n - 1) \left( \frac{n + 1}{12} \right)\)

\(D = \frac{r^2 (n - 1)(n + 1)}{12} = \frac{r^2 (n^2 - 1)}{12}\)

Conclusion:

The difference between the mean of the squares and the square of the mean is:

\(\frac{r^2(n^2 - 1)}{12}\)