Answer :

First term, a = 9, d = 12 – 9 = 3

Let, last term be an


an = a + (n – 1) d


= 9 + (n – 1)3


= 9 + 3n – 3 = 6 + 3n (i)


Let, 36th term, a36 = a + 35d


= 9 + 35 (3) = 114


Now the term is 39 more than its 36th term


an = 39 + a36


= 39 + 114 = 153


Putting the value in (i), we get


153 = 6 + 3N


3n = 153 – 6


3n = 147


n = 49


Hence, 49th term of the given A.P. is 39 more than its 36th term


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