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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
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