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