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If, for a positive integer n, the quadratic equation, \[x(x+1)+(x+1)(x+2)+.....\] \[+(x+\overline{n-1})(x+n)=10n\] has two consecutive integral solutions, then n is equal to:    JEE Main Solved Paper-2017

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            \[n=11\]