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)

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
$\therefore \:\: n - 11 = 0$
$\Rightarrow \:\: n = 11$