### The mean of $1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}, 7^{2}$ is.

A. 10 B. 20 C. 30 D. 40 Answer: Option B
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### Solution(By Apex Team)

\begin{aligned}&1^2+2^2+3^2+\ldots.+n^2=\frac{n(n+1)(2n+1)}{6}\\ &\therefore1^2+2^2+3^2+\ldots.+7^2=\left(\frac{7\times8\times15}{6}\right)=140\\ &\text{ So, required average }\\ &=\left(\frac{140}{7}\right)\\ &=20\end{aligned}

## Related Questions On Average

A. 20
B. 21
C. 28
D. 32
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A. 18
B. 20
C. 24
D. 30
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A. 10 years
B. 10.5 years
C. 11 years
D. 12 years
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### If the arithmetic mean of 0, 5, 4, 3 is a, that of -1, 0, 1, 5, 4, 3 is b and that of 5, 4, 3 is c, then the relation between a, b, and c is.

A. a = b = c
B. a : b : c = 3 : 2 : 4
C. 4a = 5b = c
D. a + b + c = 12
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