Q61 / 155 · WBJEE 2023 · Mathematics

If the \(n\) terms \(a_1, a_2, \ldots \ldots \ldots . ., a_n\) are in A.P. with increment \(r\), then the difference between the mean of their squares & the square of their mean is

  • A
    (A) \(\frac{r^2\left\{(n-1)^2-1\right\}}{12}\)
  • B
    (B) \(\frac{r^2}{12}\)
  • Correct answer
    C
    (C) \(\frac{r^2\left(n^2-1\right)}{12}\)
  • D
    (D) \(\frac{\mathrm{n}^2-1}{12}\)

Solution

Hint: \(\frac{a_1^2+a_2^2+\ldots+a_n^2}{n}-\left(\frac{a_1+a_2+\ldots+a_n}{n}\right)^2\) \(=\frac{a_1^2+\left(a_1+r\right)^2+\ldots+\left\{a_1+(n-1) r\right\}^2}{n}-\left(\frac{n a_1+r \cdot \frac{n(n-1)}{2}}{n}\right)^2\) \(\begin{aligned} & =\frac{n a_1^2+r^2 \cdot \frac{(n-1) \cdot n(2 n-1)}{6}+a_1 r(n-1) \cdot n}{n}-a_1^2-a_1 r(n-1)-\frac{r^2(n-1)^2}{4} \\ & =a_1^{2}+r^2 \frac{(n-1)(2 n-1)}{6}+a_1 r(n-1)-a_1^2-a_1 r(n-1)-\frac{r^2(n-1)^2}{4} \end{aligned}\) \(\begin{aligned} & =\frac{r^2(n-1)}{2}\left(\frac{2 n-1}{3}-\frac{n-1}{2}\right) \\ & =\frac{r^2(n-1)}{2} \cdot\left(\frac{4 n-2-3 n+3}{6}\right) \\ & =\frac{r^2(n-1)(n+1)}{12} \\ & =\frac{r^2\left(n^2-1\right)}{12}\end{aligned}\)

Hence, the correct answer is option: 'C'.

Relevant Questions