Question

For a positive integer n, if the quadratic equation, x(x + 1) + (x + 1) (x + 2) + ....+ (x + n – 1)(x + n) = 10n has two consecutive integral solutions, then n is equal to

• A. 12
• B. 9
• C. 10
• D. 11

Answer 1

The correct option is D 11

Given quadratic equation is
$x(x+1)+(x+1)(x+2)+...+(x+(n-1\left)\right)(x+n)=10n$
$\Rightarrow (x^2+x^2+....+x^2)$+[(1+3+5)+...+(2n-1)]x+[1.2+2.3+...+(n-1)n]=10n
$\Rightarrow n{x}^2+n^{2}x+\frac{n(n^2-1)}{3}-10n=0\Rightarrow 3x^2+3nx+n^2-31=0$
Let $\alpha$ and $\beta$ be the roots.
Since, $\alpha$ and $\beta$ are consecutive:
$\therefore |\alpha -\beta |=1\Rightarrow (\alpha -\beta {)}^2=1$
Again, $(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$
$\Rightarrow 1={\left(\frac{-3n}{3}\right)}^2-4\left(\frac{n^2-31}{3}\right)$
$\Rightarrow 1=n^2-\frac{4}{3}(n^2-31)\Rightarrow 3=3n^2-4n^2+124\Rightarrow n^2=121\Rightarrow n=\pm 11\therefore n=11[\because n>0]$