A) 8
B) 10
C) 15
D) 20
Correct Answer: B
Solution :
\[\Sigma {{n}^{2}}=330+\Sigma n\] \[\Rightarrow \]\[\frac{n\,(n+1)\,(2n+1)}{6}=330+\frac{n\,(n+1)}{2}\] Þ \[\frac{n\,(n+1)}{2}\left[ \frac{2n+1}{3}-1 \right]=330\] Þ \[\frac{n\,(n+1)}{2}\ .\ \frac{2(n-1)}{3}=330\] Þ \[n\,(n+1)(n-1)=990\]\[\Rightarrow \]\[n=10\].
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