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

Solution:

Given,

3^rd term of an AP = a₃ = 12

And 50^th term, a₅₀ = 106

We know that,

a_n = a + (n − 1) d

a₃ = a + (3 − 1) d

12 = a + 2d ….(1)

In the same way,

a₅₀ = a + (50 − 1) d

106 = a + 49 d ….(2)

On subtracting equation (1) from (2), we get;

94 = 47 d

d = 2 = common difference

From equation (1),

12 = a + 2(2)

a = 12 − 4 = 8

a₂₉ = a + (29 − 1) d

a₂₉ = 8 + (28)2

a₂₉ = 8 + 56 = 64

Therefore, the 29^th term is 64.