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

$\begin{array}{1 1} (1) 9 \\ (2) 10 \\ (3) 11 \\ (4) 12 \end{array} $

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