If for a positive integer n,the quadratic equ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation is $\sum_{r=1}^{n} (x + r - 1)(x + r) = 10n$.Expanding the terms: $\sum_{r=1}^{n} (x^2 + (2r - 1)x + r^2 - r) = 10n$.This simpl

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) The given equation is $\sum_{r=1}^{n} (x + r - 1)(x + r) = 10n$.Expanding the terms: $\sum_{r=1}^{n} (x^2 + (2r - 1)x + r^2 - r) = 10n$.This simplifies to $nx^2 + x \sum_{r=1}^{n} (2r - 1) + \sum_{r=1}^{n} (r^2 - r) = 10n$.Using summation formulas: $nx^2 + n^2x + \frac{(n-1)n(n+1)}{3} = 10n$.Dividing by $n$ (since $n$ is a positive integer): $x^2 + nx + \frac{n^2 - 1}{3} = 10$.$x^2 + nx + \frac{n^2 - 31}{3} = 0$.Let the two consecutive integral solutions be $\alpha$ and $\alpha + 1$.Sum of roots: $2\alpha + 1 = -n \Rightarrow \alpha = \frac{-(n+1)}{2}$.Product of roots: $\alpha(\alpha + 1) = \frac{n^2 - 31}{3}$.Substituting $\alpha$: $(\frac{-(n+1)}{2})(\frac{-(n+1)}{2} + 1) = \frac{n^2 - 31}{3}$.$(\frac{-(n+1)}{2})(\frac{1-n}{2}) = \frac{n^2 - 31}{3}$.$\frac{n^2 - 1}{4} = \frac{n^2 - 31}{3}$.$3n^2 - 3 = 4n^2 - 124$.$n^2 = 121 \Rightarrow n = 11$.

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

Similar Questions

The number of solutions of the equation $e^{\sin x} - 2e^{-\sin x} = 2$ is

Let $a$ be an integer such that all the real roots of the polynomial $2x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+10$ lie in the interval $(a, a+1)$. Then,$|a|$ is equal to ...... .

The sum of all the real numbers satisfying the equation $x^2+|x-3|=4$ is

The roots of the equation $(p - q)x^2 + (q - r)x + (r - p) = 0$ are:

The number of pairs of consecutive positive even integers such that the sum of their squares is $290$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module