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Maharashtra State BoardSSC (English Medium) 10th Standard

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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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.)

Sum
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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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