Theory Of Equations Ques 10
- For a positive integer $n$, if the quadratic equation, $x(x+1)+(x+1)(x+2)+\ldots \quad+(x+\overline{n-1})(x+n)=10 n$ has two consecutive integral solutions, then $n$ is equal to
(2017 Main)
(a) 12
(b) 9
(c) 10
(d) 11
Show Answer
Answer:
Correct Answer: 10.(d)
Solution:
- Given quadratic equation is
$x(x+1)+(x+1)(x+2)+\ldots+(x+\overline{n-1})(x+n)=10 n $
$\Rightarrow \left(x^{2}+x^{2}+\ldots+x^{2}\right)+[(1+3+5+\ldots+(2 n-1)] x $
$ +[(1 \cdot 2+2 \cdot 3+\ldots+(n-1) n]=10 n $
$ \Rightarrow n x^{2}+n^{2} x+\frac{n\left(n^{2}-1\right)}{3}-10 n=0 $
$ \Rightarrow \quad x^{2}+n x+\frac{n^{2}-1}{3}-10=0 $
$ \Rightarrow \quad 3 x^{2}+3 n x+n^{2}-31=0 $
Let $\alpha$ and $\beta$ be the roots.
Since, $\alpha$ and $\beta$ are consecutive.
$ \therefore \quad|\alpha-\beta|=1 \quad \Rightarrow \quad(\alpha-\beta)^{2}=1 $
Again, $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta$
$ \begin{array}{lcl} \Rightarrow & 1=(\frac{-3 n}{3})^{2}-4 (\frac{n^{2}-31}{3}) \\ \Rightarrow & 1=n^{2}-\frac{4}{3}\left(n^{2}-31\right) \Rightarrow 3=3 n^{2}-4 n^{2}+124 \\ \Rightarrow & n^{2}=121 \Rightarrow n= \pm 11 \\ \therefore & n=11 \quad [\because n>0] \end{array} $
BETA
