A) 5
B) 6
C) 7
D) 8
Correct Answer: C
Solution :
Let first n natural numbers are 1, 2, 3, 4, ..., n, then \[1+2+3+...+n=\frac{1}{5}[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{n}^{2}}]\] \[\Rightarrow \] \[\Sigma n=\frac{1}{5}\Sigma {{n}^{2}}\] \[\Rightarrow \] \[\frac{n(n+1)}{2}=\frac{1}{5}\left[ \frac{n(n+1)(2n+1)}{6} \right]\] \[\Rightarrow \] \[\frac{1}{2}=\frac{(2n+1)}{30}\] \[\Rightarrow \] \[2n+1=15\] \[\Rightarrow \] \[2n=14\] \[\Rightarrow \] \[n=7\]
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