Answer :

Given the sum of the 2^{nd} and 7^{th} terms of an AP is 30.

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

⇒ a_{2} = a + (2 – 1) d = a + d

⇒ a_{7} = a + (7 – 1) d = a + 6d

As given, a + d + a + 6d = 30

⇒ 2a + 7d = 30 … (1)

Also given 15^{th} term is 1 less than twice its 8^{th} term.

⇒ a_{15} = a + (15 – 1) d = a + 14d

⇒ a_{8} = a + (8 – 1) d = a + 7d

As given, a + 14d = 2 (a + 7d) – 1

⇒ a + 14d = 2a + 14d – 1

⇒ a - 1 = 0

∴ a = 1

Substituting a = 1 in (1),

⇒ 2a + 7d = 30

⇒ 2 + 7d = 30

⇒ 7d = 28

∴ d = 4

Now, we got the first term, a = 1 and the common difference, d = 4,

∴ The AP is as follows: 1, 5, 9, 13, …

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