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Find the Term of the Arithmetic Progression 9, 12, 15, 18, ... Which is 39 More than Its 36th Term. - CBSE Class 10 - Mathematics

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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

`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

  Is there an error in this question or solution?

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Solution Find the Term of the Arithmetic Progression 9, 12, 15, 18, ... Which is 39 More than Its 36th Term. Concept: Arithmetic Progression.
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