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Question
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).
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Solution
Let the three consecutive terms of an A.P. be a – d, a, a + d.
According to the first condition
a – d + a + a + d = 27
3a = 27
a = `27/3`
a = 9
According to the second condition
(a – d) × (a) × (a + d) = 504
(9 – d) × 9 × (9 + d) = 504 .....(∵ a = 9)
(9 – d) (9 + d) = `504/9`
(9 – d) (9 + d) = 56
92 – d2 = 56 .....[∵ a2 – b2 = (a – b) (a + b)]
81 – d2 = 56
81 – 56 = d2
25 = d2
Taking square root on both sides
`sqrt25 = sqrt("d"^2)`
±5 = d
a = 9, d = 5
Three consecutive terms of an A.P. are
a – d = 9 – 5 = 4
a = 9
a + d = 9 + 5 = 14
4, 9, 14
a = 9, d = –5
a – d = 9 – (–5) = 9 + 5 = 14
a = 9
a + d = 9 + 5 = 14
14, 9, 4
∴ 4, 9, 14 or 14, 9, 4
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