Question

The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms

Answer 1

Let the three consecutive terms be a – d, a and a + d

By the given first condition

a – d + a + a + d = 27

3a = 27

a = `27/3` = 9

Again by the second condition

(a – d) (a) (a + d) = 288

a(a² – d²) = 288

9(81 – d²) = 288 (a = 9)

81 – d² = `288/9`

81 – d² = 32

∴ d² = 81 – 32

= 49

d = `sqrt(49)` = ± 7

When a = 9, d = 7

a + d = 9 + 7 = 16

a = 9

a – d = 9 – 7 = 2

When a = 9, d = – 7

a + d = 9 – 7 = 2

a = 9

a – d = 9 – (–7) = 9 + 7 = 16

The three terms are 2, 9, 16 (or) 16, 9, 2