Discussion Forum

For a positive integer n, if 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

Answer:

Explanation:

Given quadratic equations is $x(x+1)+(x+1)(x+2)+.....+(x+\overline{n-1}) (x+n)=10n$

       $\Rightarrow$    $( x^{2}+x^{2}+....+x^{2})+[1+3+5+...+(2n-1)]x$

                                                       $+[1.2+2.3+...+(n-1)n]=10n$

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

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

$\Rightarrow$    3x²+3nx+n²-31=0

   Let  $\alpha$ and β be the roots

   Since,  $\alpha$ and β are consecutive

$\therefore$        | $\alpha$-β|=1

$\Rightarrow$   ( $\alpha$ -β)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= $3n^{2}-4n^{2}+124$

$\Rightarrow$   n²=121

 $\Rightarrow$   n=± 11

$\therefore$    n=11              [ n>0]