The sum of the 2nd and the 7th term of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.
Let a be the first term and d be the common difference of the AP.
Given: a2 + a7 = 30
Also, a15 = 2a8 - 1
Consider a2 + a7 = 30
⇒ (a + d) + (a + 6d) = 30
⇒ 2a + 7d = 30 ……………….. (1)
Consider a15 = 2a8 - 1
⇒ a + 14d = 2(a + 7d) - 1
⇒ a + 14d = 2a + 14d - 1
⇒ a = 1
∴ First term = a = 1
Thus, from equation (1), we get,
7d = 30 - 2a
⇒ 7d = 30 - 2
⇒ 7d = 28
⇒ d = 4
Thus, the AP is a, a + d, a + 2d, a + 3d,…
Therefore, the AP is 1, 5, 9, 13, 17,…