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JEE Main 20254 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Sequences and Series

Consider two sets A and B , each containing three numbers in A.P....

Consider two sets AA and BB, 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>0D=d+3, d \gt 0. If p+qp−q=195\frac{p+q}{p-q}=\frac{19}{5}, then p−qp-q is equal to

Options

  1. A600600
  2. B450450
  3. C630630
  4. D540540

Correct answer

D. 540540

Step-by-step solution

 Let A(a−d,a,a+d)B(b−D,b,b+D)a=12b=12p=12(144−d2)q=12(144−D2)p+qp−q=195\begin{array}{lc}\text{ Let } A(a-d, a, a+d) & B(b-D, b, b+D) \\ \quad a=12 & b=12 \\ p=12 \left(144-d^2\right) & \\ q=12 \left(144-D^2\right) & \\ \frac{p+q}{p-q}=\frac{19}{5} & \end{array}
pq=2414=127144−d2144−(d2+6d+9)=1271008−7d2=−12d2−72d+16205d2+72d−612=0d=6D=9p−q=12(D2−d3)=12(81−36)=12(45)=540 option (4)\begin{aligned} & \frac{p}{q}=\frac{24}{14}=\frac{12}{7} \\ & \frac{144-d^2}{144-\left(d^2+6 d+9\right)}=\frac{12}{7} \\ & 1008-7 d^2=-12 d^2-72 d+1620 \\ & 5 d^2+72 d-612=0 \\ & d=6 \\ & D=9 \\ & p-q=12 \left(D^2-d^3\right) \\ & =12(81-36) \\ & =12(45) \\ & =540 \\ & \text{ option }(4)\end{aligned}
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