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Which term of the A.P. −4, −1, 2, ... is 101?

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#### Solution

We have been given an arithmetic progression where

a = −4, d = −1− (−4) =3 and a_{n }= 101

We need to find which term of the given AP is 101 so, we need to find n.

Using a_{n }= a + (n − 1)d

Substituting the values in the formula we get

101 = −4 + (n−1)3

101+ 7 = 3n

3n =108

n =36

Therefore, 36th term of given A.P is 101.

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