# Find the Term of the Arithmetic Progression 9, 12, 15, 18, ... Which is 39 More than Its 36th Term. - Mathematics

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

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 + 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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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.4 | Q 32 | Page 26