Answer :

First term, a = 9, d = 12 – 9 = 3

Let, last term be a_{n}

a_{n} = a + (n – 1) d

= 9 + (n – 1)3

= 9 + 3n – 3 = 6 + 3n (i)

Let, 36^{th} term, a_{36} = a + 35d

= 9 + 35 (3) = 114

Now the term is 39 more than its 36^{th} term

a_{n} = 39 + a_{36}

= 39 + 114 = 153

Putting the value in (i), we get

153 = 6 + 3N

3n = 153 – 6

3n = 147

n = 49

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

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