JEE Main 2025 Jan 23 Shift 2, Mathematics Q25: Quadratic Equations

JEE Main2025Jan 23, Shift 2Mathematics

The roots of the quadratic equation $3 x^{2} - p x + q = 0$ are $1 0^{t h}$ and $1 1^{t h}$ terms of an arithmetic progression with common difference $\frac{3}{2}$ . If the sum of the first $11$ terms of this arithmetic progression is $88$ , then $q - 2 p$ is equal to ________.

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Solution

Use $S_{11} = \frac{11}{2} (2 a + 10 d) = 88$ , so $a + 5 d = 8$ . With $d = \frac{3}{2}$ : $a = 8 - \frac{15}{2} = \frac{1}{2} .$

The two roots:

$T_{10} = a + 9 d = \frac{1}{2} + \frac{27}{2} = 14, T_{11} = a + 10 d = \frac{1}{2} + 15 = \frac{31}{2} .$

From the quadratic $3 x^{2} - p x + q = 0$ : sum of roots $= \frac{p}{3}$ , product $= \frac{q}{3}$ .

$\frac{p}{3} = 14 + \frac{31}{2} = \frac{59}{2} \Rightarrow p = \frac{177}{2} .$

$\frac{q}{3} = 14 \cdot \frac{31}{2} = 217 \Rightarrow q = 651.$

$q - 2 p = 651 - 177 = 474.$

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q25: Quadratic Equations

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