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).

Answer 1

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