JEE Main 2025 Apr 4 Shift 2, Mathematics Q15: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Apr 4, Shift 2Mathematics

Consider two sets $A$ and $B$ , each containing three numbers in A.P. Let the sum and the product of the elements of $A$ be 36 and $p$ respectively and the sum and the product of the elements of $B$ be 36 and $q$ respectively. Let $d$ and $D$ be the common differences of AP's in $A$ and $B$ respectively such that $D = d + 3$ , $d > 0$ . If $\frac{p + q}{p - q} = \frac{19}{5}$ , then $p - q$ is equal to

A
600
B
450
C
630
D
540

Solution

Write $A = {12 - d, 12, 12 + d}$ and $B = {12 - D, 12, 12 + D}$ (each sum is $36$ ). Then

$p = 12 (144 - d^{2}), q = 12 (144 - D^{2})$ .

From $\frac{p + q}{p - q} = \frac{19}{5}$ , componendo-dividendo gives $\frac{p}{q} = \frac{24}{14} = \frac{12}{7}$ :

$\frac{144 - d ^{2}}{144 - D ^{2}} = \frac{12}{7}$ .

With $D = d + 3$ : $144 - D^{2} = 144 - d^{2} - 6 d - 9 = 135 - d^{2} - 6 d$ . So

$7 (144 - d^{2}) = 12 (135 - d^{2} - 6 d)$

$1008 - 7 d^{2} = 1620 - 12 d^{2} - 72 d$

$5 d^{2} + 72 d - 612 = 0$ .

Solving (or factoring as $5 d - stuff$ ): $d = 6$ (taking the positive root), so $D = 9$ .

Then $p - q = 12 (D^{2} - d^{2}) = 12 (81 - 36) = 12 \cdot 45 = 540$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q15: Arithmetic, Geometric And Harmonic Progressions

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