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What are the three consecutive numbers in AP whose sum is 15 and whose product is 105?

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5, 6, 8

4, 8, 9

3, 5, 7

9, 4, 5

answer is C.

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Detailed Solution

Given that the sum and product of 3 numbers are 15 and 105.
Let the consecutive numbers be

a − d, a, a + d

∴ a − d + a + a + d = 15 ⇒ 3 a = 15 ⇒ a = 5

Also given that their product is 105,

a − d × a × a + d = 105 ⇒ 5 52 − d 2 = 105 ⇒ 25 − d 2 = 21 ⇒ d 2 = 4

Taking square root on both sides,

d 2 = 4 ⇒ d = 4 ⇒ d = 2

Therefore, the required numbers are,

⇒ a − d, a, a + d ⇒ 5 − 2, 5, 5 + 2 ⇒ 3, 5, 7

Therefore, 3, 5, 7 are the required numbers.
Hence, option (3) is correct.

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