A) 11
B) 12
C) 9
D) 10
Correct Answer: A
Solution :
We have \[\sum\limits_{r=1}^{n}{(x+r-1)(x+r)=10n}\] \[\Rightarrow \]\[\sum\limits_{r=1}^{n}{({{x}^{2}}+\left| (2r-1) \right|x+({{r}^{2}}-r))}=10n\] \[\therefore \]On solving , we get
\[+\left( \frac{{{n}^{2}}-31}{3} \right)=0\] \[\therefore \]\[(2\alpha +1)=-n\Rightarrow \alpha =\frac{-(n+1)}{2}\] ? and \[\alpha (\alpha +1)=\frac{{{n}^{3}}-31}{3}\] ? \[\Rightarrow \] \[{{n}^{2}}=121\](using [a] in [b]) or \[n=11\]
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