There are two sets A and B of three numbers in A.P. whose sum is 15 and where D and d are the common differences such that $D - d = 1.$ Find the numbers if $\frac{P}{p} = \frac{7}{8}$ where P and p are products of the numbers in the two sets.

D=2, d=1

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D=5, d=3

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D=7, d=1

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D=32, d=10

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Solution

The correct option is B D=2, d=1
For the A.P having common difference D we can assume its three terms $(A - D) + A + (A + D)$

sum of the three terms is 15 so solving the equation
$(A - D) + A + (A + D) = 15$
$∴ A = 5$

For the A.P having common difference d we can assume its three terms $(A - d) + A + (A + d)$

sum of the three terms is 15 so solving the equation
$(A - d) + A + (A + d) = 15$
$∴ A = 5$

Set $A = 5 - D, 5, 5 + D$ and
set $B = 5 - d, 5, 5 + d,$

$D - d = 1$
also $\frac{P}{p} = \frac{7}{8}$

we get
$\frac{5 (25 - D^{2})}{5 (25 - d^{2})} = \frac{7}{8}$
$∴ 25 (8 - 7) = 8 {(d + 1)}^{2} - 7 d^{2}$ or

$25 = d^{2} + 16 d + 8$

so solving the above quadratic equation in d we get
$d = 1$
and $D = 2$

so terms of set B= $4, 5, 6$
terms of set A= $3, 5, 7$

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There are two sets A and B of three numbers in A.P. whose sum is 15 and where D and d are the common differences such that D−d=1. Find the numbers if Pp=78 where P and p are products of the numbers in the two sets.

There are two sets A and B of three numbers in A.P. whose sum is 15 and where D and d are the common differences such that $D - d = 1.$ Find the numbers if $\frac{P}{p} = \frac{7}{8}$ where P and p are products of the numbers in the two sets.