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← JEE Main 2017 · 2 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 62 of 90

JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 62

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

Choose one option
Answer and worked solution

Official answer: C

∑k=0n−1(x+k)(x+k+1)=nx2+n2x+(n−1)n(n+1)3=10n\displaystyle\sum_{k=0}^{n-1}(x+k)(x+k+1)=nx^2+n^2x+\frac{(n-1)n(n+1)}3=10n, i.e.
x2+nx+n2−13−10=0.x^2+nx+\frac{n^2-1}3-10=0.
Consecutive integer roots need discriminant 1: n2−43(n2−1)+40=1n^2-\frac43(n^2-1)+40=1, so n2=121n^2=121 and n=11n=11.

Topics:Roots and coefficientsSpecial series sums
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