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.)

Answer 1

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.