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

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

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Solution: 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

Or, 47d = 94

Or, 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