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. $9$
• B. $12$
• C. $11$
• D. $10$

Answer 1

The correct option is C $11$

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