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

Question

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

Tardigrade Admin · Accepted Answer

Rearranging equation, we get nx2 + 1+3+5+....+(2n-1) x + 1.2+2.3+...+(n-1)n = 10 n ⇒ nx2+n2x+((n-1)n(n+1)/3) = 10 n ⇒ x2+nx+((n2-31/3)) = 0 Given difference of roots = 1 ⇒ |α-β| = 1 ⇒ D = 1 ⇒ n2 - (4/3)(n2-31) = 1 So, n = 11

Q. If, for a positive integer n, the quadratic equation, x(x+1)+(x+1)(x+2)+....+(x+n−1​)(x+n)=10n has two consecutive integral solutions, then n is equal to :

9

10

11

12

Rearranging equation, we get nx2+{1+3+5+....+(2n−1)}x+{1.2+2.3+...+(n−1)n}=10n ⇒nx2+n2x+3(n−1)n(n+1)​=10n ⇒x2+nx+(3n2−31​)=0 Given difference of roots = 1 ⇒∣α−β∣=1 ⇒ D = 1 ⇒n2−34​(n2−31)=1 So, n=11

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