The sum of the 5^th and the 9^th terms of an AP is 30. If its 25^th term is three times its 8^th term, find the AP.

Given the sum of the 5^th and 9^th terms of an AP is 30.

We know that the nth term of an AP, a_n = a + (n – 1) d.

⇒ a₅ = a + (5 – 1) d = a + 4d

⇒ a₉ = a + (9 – 1) d = a + 8d

As given, a + 4d + a + 8d = 30

⇒ 2a + 12d = 30 … (1)

Also given 25^th term is three times its 8^th term.

⇒ a₂₅ = a + (25 – 1) d = a + 24d

⇒ a₈ = a + (8 – 1) d = a + 7d

As given, a + 24d = 3 (a + 7d)

⇒ a + 24d = 3a + 21d

⇒ -2a + 3d = 0 … (2)

Solving the (1) and (2) equations,

⇒ 15d = 30

⇒ d = 2

Substituting d = 2 in (2),

⇒ -2a + 6 = 0

⇒ 6 = 2a

⇒ a = 3

Now, we got the first term, a = 3 and the common difference, d = 2,

∴ The AP is as follows: 3, 5, 7, 9, …