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There are two sets A and B of three numbers in A.P.Whose sum i

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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 $\frac{p}{p} = \frac{7}{8}$ where P and p are products of the numbers in the two sets.

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Chloe Discussed

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.

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Text solutionVerified

(A - D) + A + (A + D) = 15

∴ A = 5

Set

A = 5 - D, 5, 5 + D

and set

B = 5 - d, 5, 5 + d,

where

D = d + 1 \frac{p}{p} = \frac{5 ( 25 - D ^{2} )}{5 ( 25 - d ^{2} )} \frac{7}{8}

∴ 25 (8 - 7) = 8 (d + 1)^{2} - 7 d^{2} \Rightarrow 25 = d^{2} + 16 d + 8

\Rightarrow d^{2} + 16 d - 17 = 0 \Rightarrow (d + 17) (d - 1) = 0

∴ d = - 17, 1

and hence

d = - 16, 2

Hence the numbers of the sets

A

and

B

are

21, 5, - 11

and

22, 5, - 12

respectively or

3, 5, 7

and

4, 5, 6

respectively.

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Question Text	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.
Answer Type	Text solution:1
Upvotes	150