Question

Find the,3positive integer in A.P such that there sum=24 and products=480

Answer 1

• Let three positive integers be

a - d, a, a + d

• And the sum of these positive integers is 24

=> a - d + a + a + d = 24

=> 3a = 24

=> a = 8 ________(eq 1)

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• Product of these (a - d, a, a + d) positive integers is 480.

=> (a - d) (a) (a + d) = 480

=> (a - d) (a + d) = $\dfrac{480}{a}$

» (x - y) (x + y) = x² - y²

=> a² - d² = $\dfrac{480}{a}$

=> (8)² - d² = $\dfrac{480}{8}$ [From (eq 1)]

=> 64 - d² = 60

=> -d² = - 4

=> d = √4

=> d = ± 2

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• We have to find A.P.

A.P. :

a - d, a, a + d

» Put a = 8 and d = + 2

=> 8 - (+2), 8, 8 + (+2)

=> 8 - 2, 8, 8 + 2

=> 6, 8, 10

» Put a = 8 and d = - 2

=> 8 - (-2), 8, 8 + (-2)

=> 8 + 2, 8, 8 - 2

=> 10, 8, 6

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A.P. =

6, 8, 10 and 10, 8, 6

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