The positive integer $n$, for which the solutions of the equation $x(x+2) + (x+2)(x+4) + \cdots + (x+2n-2)(x+2n) = \frac{8n}{3}$ are two consecutive even integers, is:
🎥 Video solution / Text Solution of this question is given below:
$x(x+2) + (x+2)(x+4) + \cdots + (x+2n-2)(x+2n) = \frac{8n}{3}$
$\Rightarrow \sum_{r=1}^{n} (x+2r-2)(x+2r) = \frac{8n}{3}$
$nx^2 + 2x\sum_{r=1}^{n}(2r-1) + 4\sum_{r=1}^{n} r(r-1) = \frac{8n}{3}$
$nx^2 + 2xn^2 + \frac{4n(n^2-1)}{3} - \frac{8n}{3} = 0$
$x^2 + 2nx + \frac{4(n^2-1)}{3} - \frac{8}{3} = 0$
Let roots be $\alpha, \beta$
$\because |\alpha - \beta| = 2$
$\Rightarrow \frac{\sqrt{D}}{|a|} = 2 \Rightarrow D = 4$
$\Rightarrow 4n^2 - 4\left(\frac{4(n^2-1)}{3} - \frac{8}{3}\right) = 4$
$\Rightarrow n^2 - \frac{4n^2}{3} = -3$
$\Rightarrow n^2 = 9$
$\Rightarrow n = 3$