If for a posiive integer n , the quadratic eq | vedclass

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Question

$\sum\limits_{r = 1}^n {(x + r - 1)(x + r) = 10n} $

$\sum\limits_{r = 1}^n {({x^2} + xr + (r - 1)x + {r^2} - r) = 10n} $

$ \Rightarrow \,\sum\li

Vedclass · Accepted Answer

$11$

Vedclass · Answer

$\sum\limits_{r = 1}^n {(x + r - 1)(x + r) = 10n} $

$\sum\limits_{r = 1}^n {({x^2} + xr + (r - 1)x + {r^2} - r) = 10n} $

$ \Rightarrow \,\sum\limits_{r = 1}^n {({x^2} + (2r - 1)x + r(r - 1) = 10n} $

$\Rightarrow n x^{2}+\{1+3+5+\ldots .+(2 n-1)\} x+\{1.2+2.3+\ldots .+(n-1)n\}$

$=10 n$

$\Rightarrow \mathrm{nx}^{2}+\mathrm{n}^{2} \mathrm{x}+\frac{(\mathrm{n}-1) \mathrm{n}(\mathrm{n}+1)}{3}=10 \mathrm{n}$

$\Rightarrow x^{2}+n x+\frac{n^{2}-31}{3}=0$

Let $\alpha$ and $\alpha+1$ be its two solutions

( $\because$ it has two consequtive integral solutions)

$\Rightarrow \alpha+(\alpha+1)=-n$

$\Rightarrow \alpha=\frac{-n-1}{2}..........(1)$

Also $\alpha(\alpha+1)=\frac{n^{2}-31}{3}........(2)$

Putting value of $(1)$ in $(2),$ we get

$=\left(\frac{n+1}{2}\right)\left(\frac{1-n}{2}\right)=\frac{n^{2}-31}{3}$

$ \Rightarrow {n^2} = 121$

$ \Rightarrow n = 11$

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

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