JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Sequences and Series

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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✓ Correct

Answer: (D)

Solution

Taking

A = (12 - d, 12, 12 + d)

and

B = (12 - D, 12, 12 + D)

from the equal sums, the products are

p = 12 (144 - d^{2})

and

q = 12 (144 - D^{2})

. The ratio

\frac{p + q}{p - q} = \frac{19}{5}

gives

\frac{p}{q} = \frac{12}{7}

, and with

D = d + 3

this leads to

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

, so

d = 6

and

D = 9

. Then

p - q = 12 (D^{2} - d^{2}) = 12 (81 - 36) = 540

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Sequences and Series

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