Verified Solution • 2025–26 Session moderate

QJEEM17Ma78c1: JEE Main 2017 Mathematics - Class 11 JEEM17Ma78c1 Solution: If for a positive integer n, the quadratic equation has two…

If for a positive integer n, the quadratic equation

x(x+1)+(x+1)(x+2)x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) +....+(x+n−1‾)(x+n)+ .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right) =10n= 10n

has two consecutive integral solutions, then n is equal to :

Multiple Choice Options

  • A 9
  • B 10
  • C 11
  • D 12

Step-by-Step Verified Solution

∑r=1n(x+r−1)(x+r)=10n\sum\limits_{r = 1}^n {\left( {x + r - 1} \right)\left( {x + r} \right)} = 10n

⇒\Rightarrow ∑r=1n(x2+xr+(r−1)x+r2−r)=10n\sum\limits_{r = 1}^n {\left( {{x^2} + xr + \left( {r - 1} \right)x + {r^2} - r} \right)} = 10n

⇒\Rightarrow ∑r=1n(x2+(2r−1)x+r(r−1))=10n\sum\limits_{r = 1}^n {\left( {{x^2} + \left( {2r - 1} \right)x + r\left( {r - 1} \right)} \right)} = 10n

⇒\Rightarrow nx2+{1+3+5+....+(2n−1)}xn{x^2} + \left\{ {1 + 3 + 5 + .... + \left( {2n - 1} \right)} \right\}x

+{1.2+2.3+...+(n−1)n}+ \left\{ {1.2 + 2.3 + ... + \left( {n - 1} \right)n} \right\} = 10n

⇒\Rightarrow nx2+n2x+n(n2−1)3=10nn{x^2} + {n^2}x + {{n\left( {{n^2} - 1} \right)} \over 3} = 10n

⇒\Rightarrow x2+nx+(n2−31)3=0{x^2} + nx + {{\left( {{n^2} - 31} \right)} \over 3} = 0

Let α\alpha and α\alpha + 1 be its two solutions

∴\therefore α\alpha + ( α\alpha + 1) = -n

⇒\Rightarrow α\alpha = −n−12{{ - n - 1} \over 2} ....(1)

Also α\alpha ( α\alpha + 1) = (n2−31)3{{\left( {{n^2} - 31} \right)} \over 3} ......(2)

Putting value of (1) in (2), we get

−(n+12)(1−n2)=(n2−31)3- \left( {{{n + 1} \over 2}} \right)\left( {{{1 - n} \over 2}} \right) = {{\left( {{n^2} - 31} \right)} \over 3}

⇒\Rightarrow n2 = 121

⇒\Rightarrow n = 11

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