JEE Main 2026 Apr 6 Shift 2, Mathematics Q2: Summation of Series

JEE Main2026Apr 6, Shift 2Mathematics

Consider the quadratic equation $(n^{2} - 2 n + 2) x^{2} - 3 x + (n^{2} - 2 n + 2)^{2} = 0$ , $n \in R$ . Let $α$ be the minimum value of the product of its roots and $β$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is $α$ and the common ratio is $\frac{α}{β}$ , is:

A
$\frac{61}{37}$
B
$\frac{121}{81}$
C
$\frac{364}{243}$
D
$\frac{1093}{729}$

Solution

Let $A = n^{2} - 2 n + 2 = (n - 1)^{2} + 1 \geq 1$ .

Product of roots $= \frac{A ^{2}}{A} = A$ . Minimum at $n = 1$ : $α = 1$ .

Sum of roots $= \frac{3}{A} = \frac{3}{( n - 1 ) ^{2} + 1}$ . Maximum at $n = 1$ : $β = 3$ .

Common ratio $r = \frac{α}{β} = \frac{1}{3}$ , first term $= 1$ , six terms:

$S_{6} = \frac{1 - ( 1/3 ) ^{6}}{1 - 1/3} = \frac{3}{2} [1 - \frac{1}{729}] = \frac{3}{2} \cdot \frac{728}{729} = \frac{364}{243}$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q2: Summation of Series

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