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Question -

Find the term of the arithmetic progression 9, 12, 15, 18, … which is 39 more than its 36th term (C.B.S.E. 2006C)



Answer -

In the given A.R 9,12, 15, 18, …
First term (a) = 9
and common difference (d) = 12 – 9 = 3
and an = a + (n – 1) d
Now a36 = a + (36 – 1) d = 9 + 35 x 3 = 9 + 105 = 114
Let the an be the required term
an = a + (n – 1) d
= 9 + (n – 1) x 3 = 9 + 3n – 3 = 6 + 3n
But their difference is 39
an – a36 = 39
=> 6 + 3n – 114 = 39
=> 114 – 6 + 39 = 3n
=> 3n = 147
=> n = 49
Required term is 49th

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