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

JEE Main · 2017 · 02 Apr 2017 · Offline · Quadratic Equations and Inequalities

If, for a positive integer n , the quadratic equation,

x (x + 1) + (x + 1) (x + 2) + \dots

+ (x + \overline{n - 1}) (x + n) = 10 n

has two consecutive integral solutions, then n is equal to :

Answer: (A)

Each term is

(x + k) (x + k + 1)

. Summing from

k = 0

n - 1

and dividing through by

n

reduces the equation to

x^{2} + n x + \frac{( n - 1 ) ( n + 1 )}{3} = 10

. Two consecutive integer roots differ by

1

, so their discriminant must equal

1

. Testing shows

n = 11

gives

x^{2} + 11 x + 30 = 0

with roots

- 5

and

- 6

, which are consecutive integers. Hence

n = 11

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option