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Question
In an A.P. sum of three consecutive terms is 27 and their products is 504. Find the terms. (Assume that three consecutive terms in an A.P. are a – d, a, a + d.)
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Solution
Let the three consecutive terms in an A.P. be a – d, a, and a + d.
According to the first condition,
sum of three consecutive terms is 27.
a – d + a + a + d = 27
∴ 3a = 27
∴ a = `27/3`
∴ a = 9 ..........(i)
According to the second condition,
the product of the three numbers is 504.
(a – d) a (a + d) = 504
∴ a(a2 – d2) = 504
∴ 9(92 – d2) = 504 ........[From(i)]
∴ 81 – d2 = `504/9`
∴ 81 – d2 = 56
∴ d2 = 81 – 56
∴ d2 = 25
Taking square root of both sides, we get
d = ±5
When d = 5 and a = 9,
a – d = 9 – 5 = 4
a = 9
a + d = 9 + 5 = 14
When d = – 5 and a = 9,
a – d = 9 – (– 5) = 9 + 5 = 14
a = 9
a + d = 9 – 5 = 4
∴ The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.
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