# Find the term of the arithmetic progression $9, 12, 15, 18, â€¦$ which is 39 more than its 36th term.

Given:

Given A.P. is $9, 12, 15, 18, …$

To do:

We have to find the term of the given A.P. which is 39 more than its 36th term.

Solution:

Here,

$a_1=9, a_2=12, a_3=15$

Common difference $d=a_2-a_1=12-9=3$

We know that,

nth term $a_n=a+(n-1)d$

Therefore,

$a_{36}=9+(36-1)(3)$

$=9+105$

$=114$

39 more than the 36th term $=39+114=153$

This implies,

$a_{n}=9+(n-1)3$

$153=9+3n-3$

$3n=153-6$

$n=\frac{147}{3}$

$n=49$

Hence, 49th term is 39 more than the 36th term.â€Š

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