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Question
Find three numbers in A.P. whose sum is 24 and whose product is 440.
Solution
Let the three numbers in A.P. be a – d, a and a + d.
∴ (a – d) + a + (a + d) = 24
`=>` 3a = 24
`=>` a = 8 ...(1)
Also, (a – d) × a × (a + d) = 440
`=>` (a2 – d2) × a = 440
`=>` (82 – d2) × 8 = 440 ...[From (1)]
`=>` 64 – d2 = 55
`=>` d2 = 9
`=>` d = ±3
When a = 8 and d = 3
Required terms = a – d, a and a + d
= 8 – 3, 8, 8 + 3
= 5, 8, 11
When a = 8 and d = –3
Required term = a – d, a and a + d
= 8 – (–3), 8, 8 + (–3)
= 11, 8, 5
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