InterviewSolution
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InterviewSolution
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If, for a positive integer `n ,`the quadratic equation,`x(x+1)+(x-1)(x+2)++(x+ n-1)(x+n)=10 n`has two consecutive integral solutions, then `n`is equal to :`10`(2) `11`(3) `12`(4) `9`A. 11B. 12C. 9D. 10 |
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Answer» Correct Answer - A 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` |
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