In an A.P. sum of three consecutive terms is 27 and their product is 5

1.	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» Let the three consecutive terms in an A.P. be a – d, a and a + d. According to the first condition, a – d + a + a + d = 27 ∴ 3a = 27 ∴ a = 27/3 ∴ a = 9 ….(i) According to the second condition, (a – d) a (a + d) = 504 ∴ a(a² – d² ) = 504 ∴ 9(a² – d²) = 504 …[From (i)] ∴ 9(81 – d² ) = 504 ∴ 81 – d² = 504 ∴ 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.

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»

Let the three consecutive terms in an A.P. be

a – d, a and a + d.

According to the first condition,

a – d + a + a + d = 27

∴ 3a = 27

∴ a = 27/3

∴ a = 9 ….(i)

According to the second condition,

(a – d) a (a + d) = 504

∴ a(a² – d² ) = 504

∴ 9(a² – d²) = 504 …[From (i)]

∴ 9(81 – d² ) = 504

∴ 81 – d² = 504

∴ 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.