Answer :

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.

⇒ a_{5} = a + (5 – 1) d = a + 4d

⇒ a_{9} = 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.

⇒ a_{25} = a + (25 – 1) d = a + 24d

⇒ a_{8} = 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, …

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