Find the term of the arithmetic progression 9, 12, 15, 18, ... which is

Question:

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

Solution:

In the given problem, let us first find the 36st term of the given A.P.

A.P. is 9, 12, 15, 18 …

Here,

First term (a) = 9

Common difference of the A.P. $(d)=12-9=3$

Now, as we know,

$a_{n}=a+(n-1) d$

So, for 36th term (n = 36),

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

$=9+35(3)$

$=9+105$

 

$=114$

Let us take the term which is 39 more than the 36th term as an. So,

$a_{n}=39+a_{36}$

$=39+114$

$=153$

Also, $a_{n}=a+(n-1) d$

$153=9+(n-1) 3$

$153=9+3 n-3$

$153=6+3 n$

$153-6=3 n$

Further simplifying, we get,

$147=3 n$

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

$n=49$

Therefore, the $49^{\text {th }}$ term of the given A.P. is 39 more than the $36^{\text {th }}$ term

 

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