If, for a positive integer `n ,` the quadratic equation, `x(x+1)+(x-1)(x+2)++(x+ n-1)(x+n)=10 n` has two consecutive integral solutions, then `n` is equal to : ` (1) 10` (2) `11` (3) `12` (4) `9`

Options

A. (a) `11`
B. (b) `12`
C. (c) `9`
D. (d) `10`

Answer

A

Explanation

`:'x(x+1)+(x+1)(x+2)+……..+(x+bar(n-1))(x+n)=10n`
`impliesnx^(2)+x(1+3+5+….+(2n-1))+(1.2+2.3)+……..+(n-1).n)=10n`
or `nx^(2)+n^(2)x+1/3(n-1)n(n+1)=10n`
or `3x^(2)+3nx+(n^(2)-1)=30( :' n!=0`
or `3x^(2)+3nx+(n^(2)-31)=0`
`:'|alpha-beta|=1`
or `(alpha-beta)^(2)=1`
or `D/(a^(2))=1`
or `D=a^(2)`
or `9n^(2)-12.(n^(2)-31)=9`
or `n^(2)=121`
`:.n=11`