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

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Solution

Q.
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 [2025]

1 450

2 630

3 540

4 600

Ans.
(3)

Let A = (a – d, a, a + d) and B = (b – D, b, b + D)

For set A, Sum = 36, product = p

For set B, Sum = 36, product = q

$\Rightarrow a = 12, p = 12 (144 - d^{2})$

$\Rightarrow b = 12, q = 12 (144 - D^{2})$

$= 12 (144 - {(d + 3)}^{2}) = 12 (144 - d^{2} - 6 d - 9)$

Now, $\frac{p + q}{p - q} = \frac{19}{5} \Rightarrow \frac{p}{q} = \frac{12}{7}$

$\Rightarrow \frac{144 - d^{2}}{144 - d^{2} - 6 d - 9} = \frac{12}{7}$

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

$\Rightarrow d = \frac{- 72 \pm \sqrt{{(132)}^{2}}}{10}$

$\Rightarrow d = 6 and D = 9$ [ $∵ d > 0$ ]

$∴ p - q = 12 (D^{2} - d^{2}) = 12 (9^{2} - 6^{2}) = 540$ .

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