If for a positive integer n,the quadratic equ | vedclass

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(A) The given equation is $\sum_{r=1}^{n} (x + r - 1)(x + r) = 10n$.Expanding the terms: $\sum_{r=1}^{n} (x^2 + (2r - 1)x + r^2 - r) = 10n$.Summing ov

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$11$

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(A) The given equation is $\sum_{r=1}^{n} (x + r - 1)(x + r) = 10n$.Expanding the terms: $\sum_{r=1}^{n} (x^2 + (2r - 1)x + r^2 - r) = 10n$.Summing over $r$: $nx^2 + x \sum_{r=1}^{n} (2r - 1) + \sum_{r=1}^{n} (r^2 - r) = 10n$.Using summation formulas: $\sum_{r=1}^{n} (2r - 1) = n^2$ and $\sum_{r=1}^{n} (r^2 - r) = \frac{(n-1)n(n+1)}{3}$.So,$nx^2 + n^2x + \frac{n(n^2 - 1)}{3} = 10n$.Dividing by $n$ $(n 
eq 0)$: $x^2 + nx + \frac{n^2 - 1}{3} = 10$.$x^2 + nx + \frac{n^2 - 31}{3} = 0$.Let the roots be $\alpha$ and $\alpha + 1$. Sum of roots: $2\alpha + 1 = -n \Rightarrow \alpha = \frac{-(n+1)}{2}$.Product of roots: $\alpha(\alpha + 1) = \frac{n^2 - 31}{3}$.Substituting $\alpha$: $\left(\frac{-(n+1)}{2}ight) \left(\frac{-(n+1)}{2} + 1ight) = \frac{n^2 - 31}{3}$.$\left(\frac{-(n+1)}{2}ight) \left(\frac{1-n}{2}ight) = \frac{n^2 - 31}{3} \Rightarrow \frac{n^2 - 1}{4} = \frac{n^2 - 31}{3}$.$3n^2 - 3 = 4n^2 - 124 \Rightarrow n^2 = 121 \Rightarrow n = 11$.

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

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