Question

Which term of the A.P 9, 12, 15, 18, .... is 39 more than its 36^th term?

Answer 1

Given A.P. = 9, 12, 15, 18...........

a = 9

d = 12 - 9

d = 3

Now, according to question.

a₃₆ = a + (n - 1)d

a₃₆ = 9 + (36 - 1)3

= 9 + 35*3

= 9 + 105

a₃₆ = 114

The required term is 39 more than its 36th term.

So,

⇒ a₃₆ + 39

⇒ 114 + 39 = 153

an = 153

153 = a + (n - 1)d

153 = 9 + (n - 1)3

153 = 9 + 3n - 3

3n = 153 + 3 - 9

3n = 147

n = 147/3

n = 49

So, the required term is 49th term of the given A.P.