An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29thterm.

Given, a3 = 12 and a50 = 106


a3 = a + 2d = 12


a50 = a + 49d = 106


Subtracting 3rd term from 50th term, we get;


a + 49d – a – 2d = 106 – 12


47d = 94


d = 2


Substituting the value of d in 12th term, we get;


a + 2 x 2 = 12


Or, a + 4 = 12


Or, a = 8


Now, 29th term can be calculated as follows:


a29 = a + 28d


= 8 + 28 x 2


= 8 + 56 = 64


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