Q22 of 50 Page 1

The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.

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


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


⇒ a2 = a + (2 – 1) d = a + d


⇒ a7 = a + (7 – 1) d = a + 6d


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


⇒ 2a + 7d = 30 … (1)


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


⇒ a15 = a + (15 – 1) d = a + 14d


⇒ a8 = 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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