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. 12
• B. 9
• C. 10
• D. 11

Answer 1

Answer: (D)

11

Given,

$x(x+1)+(x+1)(x+2)+.....(x+n-1)(x+n)=10n$ has two consecutive integral solutions.

$x^2+x+x^2+(1+2)x+1\cdot 2+\cdots$

$+x^2+(n+1+n)x+n\cdot n-1=10n$

$\Rightarrow n\cdot x^2+1\cdot x+(1+2)x+(2+3)x....+(2n-1)x$

$+1\cdot 2+6++...(5-1)\cdot n-10n=0$

$\Rightarrow n{x}^2+x(1+3+5+\cdots (2n-1)+\frac{n(n^2-1)}{3}-10n=0$

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

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

A/Q the Root $(\alpha ,\beta )$ of the above equation is consecutive.

$\alpha +\beta =n$ or $2-\beta =1\alpha >\beta$

$\alpha \beta =\frac{n^2-1-30}{3}$

$\begin{array}{cc}\therefore \alpha -\beta =\frac{\sqrt{b}}{a}& [\because a=1]\end{array}$

$\begin{array}{cc}2-\beta =\sqrt{D}& \text{(squaring both sides)}\\ (2-\beta {)}^2=D& \end{array}$

$\begin{array}{cc}& (\alpha +\beta {)}^2-42\beta =1\\ \Rightarrow & n^2-4(\frac{n^2-1}{3}-10)=1\\ \Rightarrow & 3n^2-4n^2+4+120-3=0\\ \Rightarrow & n^2=121\\ \Rightarrow & n=11\because n\ne -11\\ & \therefore \text{(option}D=11)\end{array}$