Sequences & Series
Product Telescoping — Consecutive Even Integer Roots
nta_pyq_2026_jan
Grade 11
Question:
The positive integer $n$, for which the solutions of the equation $x(x+2)+(x+2)(x+4)+\cdots+(x+2n-2)(x+2n)=\dfrac{8n}{3}$ are two consecutive even integers, is:
Step-by-Step Solution
Key Concept: Sum $=nx^2+2n^2x+\tfrac{4(n^2-1)}{3}=\tfrac{8n}{3}$. Dividing by $n$: $x^2+2nx+\tfrac{4(n^2-1)}{3}-\tfrac{8}{3}=0$. By Vieta's: $x_1+x_2=-2n$, $x_1x_2=\tfrac{4n^2-12}{3}$.
$n=3$.
Correct Answer: 1