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:

a1, a1 + r, a1 + 2r, ..., a1 + (n-1)r

The mean of these terms is the sum of all terms divided by the total number of terms:

Mean = (a1 + a1 + r + a1 + 2r + ... + a1 + (n-1)r) / n

The mean of their squares is:

Mean of Squares = [(a1)² + (a1 + r)² + (a1 + 2r)² + ... + (a1 + (n-1)r)²] / n

Now, let's find the difference between the mean of their squares and the square of their mean:

Difference = [(a1)² + (a1 + r)² + (a1 + 2r)² + ... + (a1 + (n-1)r)²] / n - [(a1 + a1 + r + a1 + 2r + ... + a1 + (n-1)r) / n]²

Now, the key observation is that the terms (a1 + r)², (a1 + 2r)², ..., (a1 + (n-1)r)² are in an arithmetic progression with a common difference of 2r.

Using the formula for the sum of squares of an A.P., the sum of these squares is given by:

Sum of Squares of A.P. = [n * ((2r(n-1)(2r(n-1) + 1) / 6)] + (n-1)r²

Similarly, the sum of the A.P. a1, a1 + r, a1 + 2r, ..., a1 + (n-1)r is given by:

Sum of A.P. = [n * (a1 + (a1 + (n-1)r)) / 2]

Now, if we substitute these values back into the "Difference" expression and simplify, we'll find that the answer is indeed 12r²(n² - 1).

the answer is derived by analyzing the properties of arithmetic progressions, using the sum of squares formula for an A.P., and manipulating the expressions to arrive at the given result of 12r²(n² - 1).

The correct answer is option (C): \(\frac{r^2(n^2-1)}{12}\)