If the mean of n observations 12,22,32.....n2 is 46 n11 then n is equal to

∵12+22+32+...+n2n=46n11⇒n(n+1)(2n+1)6n=46n11⇒11(2n2+3n+1)=276n⇒22n2−243n+11=0⇒(n−11)(22n−1)=0∴n=11

If the mean of n observations 12,22,32.....n2 is 46 n11 then n is equal to

∵12+22+32+...+n2n=46n11⇒n(n+1)(2n+1)6n=46n11⇒11(2n2+3n+1)=276n⇒22n2−243n+11=0⇒(n−11)(22n−1)=0∴n=11

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

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$∵ \frac{1^{2} + 2^{2} + 3^{2} + ... + n^{2}}{n} = \frac{46 n}{11}$

$\Rightarrow \frac{n (n + 1) (2 n + 1)}{6 n} = \frac{46 n}{11}$

$\Rightarrow 11 (2 n^{2} + 3 n + 1) = 276 n$

$\Rightarrow 22 n^{2} - 243 n + 11 = 0 \Rightarrow (n - 11) (22 n - 1) = 0$

$∴ n = 11$

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If the mean of n observations 12,22,32.....n2 is 46 n11 then n is equal to

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