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

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#### Solution

In the given problem, let us first find the 36^{st} 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

`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 36^{th} term as *a*_{n}. So,

`a_n = 39 + a_36`

= 39 + 114

= 153

Also `a_n = a + (n -1)d`

153 = 9 + (n -1)3

153 = 9 + 3n - 3

153 = 6 + 3n

153 - 6 = 3n

Further simplifying, we get,

147 = 3n

`n = 147/3`

n = 49

Therefore, the 49 th term if the given A.P. is 39 more than the 36 th term

Concept: Arithmetic Progression

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