Answer :

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,…


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