Question

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

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

Answer: (C)

Rearranging equation, we get
$nx^{2} + \left\{1+3+5+....+\left(2n-1\right)\right\}x + \left\{1.2+2.3+...+\left(n-1\right)n\right\} = 10\,n$
$\Rightarrow nx^{2}+n^{2}x+\frac{\left(n-1\right)n\left(n+1\right)}{3} = 10\,n$
$\Rightarrow x^{2}+nx+\left(\frac{n^{2}-31}{3}\right) = 0$
Given difference of roots = 1
$\Rightarrow \left|\alpha-\beta\right| = 1$
$\Rightarrow $ D = 1
$\Rightarrow n^{2} - \frac{4}{3}\left(n^{2}-31\right) = 1$
So, $n = 11$