Question

If the mean of n observations $1^2, 2^2, 3^2....,n^2$ is $\frac{46n}{11}$ , then n is equal to

• A. 11
• B. 12
• C. 23
• D. 22

Answer 1

Answer: (A)

11

Answer 2

Mean of n observations is ${\frac{1^2 + 2^2 + 3^2 + .... n^2}{n} = \frac{n(n + 1)(2n + 1)}{6n}}$ From the description of the problem: ${\frac{(n + 1)(2n +1)}{6} =\frac{46n}{11}}$ $\Rightarrow \:\:\: {11 \times (2n^{2} + 3n + 1) = 6 \times 46 n}$ $\Rightarrow \:\:\: { 22n^{2} + 33 n + 11 = 276 n}$ $\Rightarrow \:\:\: { 22n^{2} + 243 n + 11 = 0}$ $\Rightarrow \:\:\: { 22n^{2} + 242 n - n + 11 = 0}$ $\Rightarrow \:\:\: { 22n(n - 11 ) - 1 (n -11) = 0}$ $\Rightarrow \:\:\: {(n - 11) (22n - 1) = 0}$ Now, ${22n - 1 = 0}$ $\Rightarrow \:\: { n = \frac{1}{22} }$ which is discarded as n cannot be a fraction $\therefore \:\: { n - 11 = 0}$ $\Rightarrow \:\: { n = 11}$