Question

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

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The correct option is C

11

Step 1: Rearranging the given equation

$\mathrm{x}(\mathrm{x}+1)+(\mathrm{x}+1)(\mathrm{x}+2)+....+(\mathrm{x}+\overline{\mathrm{n}-1})(\mathrm{x}+\mathrm{n})=10\mathrm{n}$ can be written into general form as,

$\sum _{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{x}+\mathrm{r}-1)(\mathrm{x}+\mathrm{r})=10\mathrm{n}$

$$\begin{gathered} \sum _{\mathrm{r}=1}^{\mathrm{n}}{\mathrm{x}}^2+\mathrm{rx}-\mathrm{x}+\mathrm{rx}+{\mathrm{r}}^2-\mathrm{r}=10\mathrm{n} \\ \sum _{r=1}^n{x}^2+2rx-x+r^2-r=10n \end{gathered}$$

for n^th term, this can be written as,

$$\begin{gathered} {\mathrm{nx}}^2+\mathrm{x}\sum _{\mathrm{r}=1}^{\mathrm{n}}(2\mathrm{r}-1)+\sum _{\mathrm{r}=1}^{\mathrm{n}}({\mathrm{r}}^2-\mathrm{r})=10\mathrm{n} \\ {\mathrm{nx}}^2+\mathrm{x}({\mathrm{n}}^2+\mathrm{n}-\mathrm{n})+\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}-\frac{\mathrm{n}(\mathrm{n}+1)}{2}=10\mathrm{n} \\ {\mathrm{nx}}^2+{\mathrm{n}}^2\mathrm{x}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{(2\mathrm{n}+1)}{3}-1\right)-10\mathrm{n}=0 \\ {\mathrm{nx}}^2+{\mathrm{n}}^2\mathrm{x}+\mathrm{n}\frac{(\mathrm{n}+1)}{2}\left(\frac{2\mathrm{n}+1-3}{3}\right)-10\mathrm{n}=0 \\ {\mathrm{nx}}^2+{\mathrm{n}}^2\mathrm{x}+\mathrm{n}\frac{(\mathrm{n}+1)}{2}\left(\frac{2\mathrm{n}-2}{3}\right)-10\mathrm{n}=0 \\ {\mathrm{nx}}^2+{\mathrm{n}}^2\mathrm{x}+\mathrm{n}(\mathrm{n}+1)\left(\frac{\mathrm{n}-1}{3}\right)-10\mathrm{n}=0 \\ {\mathrm{nx}}^2+{\mathrm{n}}^2\mathrm{x}+({\mathrm{n}}^2+n)\left(\frac{\mathrm{n}-1}{3}\right)-10\mathrm{n}=0 \\ {\mathrm{nx}}^2+{\mathrm{n}}^2\mathrm{x}+\mathrm{n}\left(\frac{{\mathrm{n}}^2-31}{3}\right)=0 \\ \mathrm{n}\left[{\mathrm{x}}^2+\mathrm{nx}+\frac{{\mathrm{n}}^2-31}{3}\right]=0 \end{gathered}$$

step 2: Using $\alpha$ and $\beta$ as the two consecutive roots of the derived equation

Given, difference of two roots is 1

$\left|\alpha -\beta \right|=1$

let $\alpha$ and $\beta$ be the roots of the equation, then

$\begin{array}{rcl}\mathrm{\alpha }+\mathrm{\beta }& =& -\mathrm{n}\\ \mathrm{\alpha \beta }& =& \frac{{\mathrm{n}}^2-31}{3}\\ {\left(\mathrm{\alpha }-\mathrm{\beta }\right)}^2& =& 1\end{array}$

$\begin{array}{rcl}& \Rightarrow & {\left(\alpha +\beta \right)}^2-4\alpha \beta =1\\ & \Rightarrow & n^2-4\left(\frac{n^2-31}{3}\right)-1=0\\ & \Rightarrow & n^2+\frac{-4n^2+124-3}{3}=0\\ & \Rightarrow & \frac{3n^2-4n^2+121}{3}=0\\ & \Rightarrow & -n^2+121=0\\ & \Rightarrow & -n^2=-121\\ & \Rightarrow & n=11\end{array}$

Hence, the correct answer is option (c).