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

a₃ = 12 and a_n = 106

Find: a₂₉

a₃  = a + (3 - 1)d = a + 2d

a + 2d = 12 .... (1)

Thus, 50^th term =106 [Since, n = 50]

a + (50 - 1)d = 106

a + 49d = 106........ (2)

By solving equations (1) and (2) we get,

a + 49d - (a + 2d) = 106 - 12

47d = 94

d = 2

Substitute d = 2 in equation (1)

a + 2 × 2 = 12

a + 4 = 12

a = 12 - 4

a = 8

 a₂₉ = a + (29 - 1)d

a₂₉ = 8 + (28) 2

a₂₉ = 8 + 56

a₂₉ = 64

Thus, 29^th term of the AP is 64.