Question

If, for a positive integer n, the quadratic equation, $x(x+1)+(x+1)(x+2)+\dots +(x+\frac{1}{n-1})(x+n)=10n$ has two consecutive integral solutions, then n is equal to

• A. 11
• B. 12
• C. 9
• D. 10

Answer 1

The correct option is A 11

$n{x}^2+x(1+3+5+...+(2n-1))+(1.2+2.3+...+(n-1).n)-10n=0$

$\Rightarrow n{x}^2+x{n}^2+\frac{n(n^2-1)}{3}-10n=0$

$\Rightarrow x^2+nx+\frac{(n^2-1)}{3}-10=0$

Let $\alpha$ and $\beta$ be the roots of the equation then $\alpha +\beta =-n$ and $\alpha \beta =\frac{(n^2-1)}{3}-10$

And ${(\alpha -\beta )}^2=1$

$\Rightarrow {(\alpha +\beta )}^2-4\alpha \beta =1$

$\Rightarrow n^2-4(\frac{(n^2-1)}{3}-10)=1$

$\Rightarrow 3n^2-4n^2+4+120-3=0$

$\Rightarrow n^2=121$

$\therefore n=11$