A) 11
B) 12
C) 13
D) 14
Correct Answer: B
Solution :
Given, the sum of first n natural numbers \[=\frac{1}{78}\] (the sum of their cubes) \[\Rightarrow \] \[\frac{n(n+1)}{2}=\frac{1}{78}\times \frac{{{n}^{2}}{{(n+1)}^{2}}}{4}\] \[\Rightarrow \] \[156=n(n+1)\] \[\Rightarrow \] \[{{n}^{2}}+n-156=0\] \[\Rightarrow \] \[(n+13)(n-12)=0\] \[\Rightarrow \] \[n=12\] \[(\because \,n\ne -13)\]You need to login to perform this action.
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