JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(D=d+3, d \gt 0\). If \(\frac{p+q}{p-q}=\frac{19}{5}\), then \(p-q\) is equal to
- A \(600\)
- B \(450\)
- C \(630\)
- D \(540\)
Answer & Solution
Correct Answer
(D) \(540\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{lc}\text { Let } A(a-d, a, a+d) & B(b-D, b, b+D) \\ \quad a=12 & b=12 \\ p=12\left(144-d^2\right) & \\ q=12\left(144-D^2\right) & \\ \frac{p+q}{p-q}=\frac{19}{5} & \end{array}\)…
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