JEE Main 2026 — Mathematics Algebra

The given quadratic equation is $(n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0$(n2−2n+2)x2−3x+(n2−2n+2)2=0.

Let $k = n^2 - 2n + 2 = (n-1)^2 + 1$k=n2−2n+2=(n−1)2+1. Since $(n-1)^2 \ge 0$(n−1)2≥0 for all $n \in \mathbb{R}$n∈R, the minimum value of $k$k is $1$1 at $n=1$n=1.

The product of the roots is given by $P = \dfrac{(n^2 - 2n + 2)^2}{n^2 - 2n + 2} = n^2 - 2n + 2 = k$P=n2−2n+2(n2−2n+2)2​=n2−2n+2=k.

The minimum value of the product of the roots is $\alpha = \min(k) = 1$α=min(k)=1.

The sum of the roots is given by $S = \dfrac{3}{n^2 - 2n + 2} = \dfrac{3}{k}$S=n2−2n+23​=k3​.

The maximum value of the sum of the roots occurs when $k$k is minimum. Thus, $\beta = \max\left(\dfrac{3}{k}\right) = \dfrac{3}{1} = 3$β=max(k3​)=13​=3.

We are given a Geometric Progression (G.P.) whose first term is $a = \alpha = 1$a=α=1 and common ratio is $r = \dfrac{\alpha}{\beta} = \dfrac{1}{3}$r=βα​=31​.

The sum of the first six terms of this G.P. is:

$$S_6 = a \dfrac{1 - r^6}{1 - r} = 1 \cdot \dfrac{1 - \left(\dfrac{1}{3}\right)^6}{1 - \dfrac{1}{3}}$$S6​=a1−r1−r6​=1⋅1−31​1−(31​)6​

$$S_6 = \dfrac{1 - \dfrac{1}{729}}{\dfrac{2}{3}} = \dfrac{\dfrac{728}{729}}{\dfrac{2}{3}}$$S6​=32​1−7291​​=32​729728​​

$$S_6 = \dfrac{728}{729} \times \dfrac{3}{2} = \dfrac{364}{243}$$S6​=729728​×23​=243364​

Answer: $\dfrac{364}{243}$243364​