For a positive integer n, if the quadratic equation, equation, `x(x+1)+(x+1)(x+2)+...+(x+n-1)(x+n)=10n` has two consective integral solutions, then n is equal to

Options

A. 11
B. 12
C. 9
D. 10

Answer

A

Explanation

we have
`x(x+1)+(x+1)(x+2)+…+(x+bar(n-1))(x+n)=10n `
Roots of above equation are consecutive integers .
Let roots be `alpha " and " alpha +1`
`rArr alpha(alpha +1)+(alpha +1)(alpha+2)+…..+(alpha +n )(alpha +n+1)=10 n `
and `(alpha +1)(alpha+2)+(alpha+2)(alpha+3)+.....+(alpha +n)(alpha+n+1)=10n `
Subractiong (2) from (1) we get
`alpha(alpha+1)-(alpha +n)(alpha +n+1)=0`
`rArr alph^2+alpha-alpha^2(2n +1)alpha-n(n+1)=0`
`rArr alpha = (n+1)/(2)`
Putting this vlaue in the original equation i.e
`nx^2+x underset(r=1)overset(n)Sigma (2r-1)+underset(r=1)overset(n)Sigma(r-1)r= 10n` ,we get `rArr 3(n+1)^2-6n(n+1)+2(n+1)-6(n+1)=120`
`rArr n^2=121`
`rArr n=11`