### JEE Main Solved Paper 2017 Question 2

##### Question: 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]

#### Options:

A) 11

B) 12

C) 9

D) 10

## Show Answer

#### Answer:

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}+| (2r-1) |x+(r^{2}-r))}=10n $

$ \therefore $ On solving , we get $ +( \frac{n^{2}-31}{3} )=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 $