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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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प्रश्न

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.)

योग
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उत्तर

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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