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

Here, a₃ = 12 and a_50 = 106. To find : a_29
Given that: a₃ = a + (3 – 1)d = 12
⇒ a + 2d = 12
⇒ a = 12 – 2d . . .(1)
and a_50 = 106
⇒ a + 49d = 106
Putting the value of a from equation (1), we get
12 – 2d + 49d = 106
⇒ 47d = 94 ⇒ d = 2
putting the value of d in equation (1), we get
a = 12 – 2(2) = 8
Therefore, a₂₉ = a + 28d = 8 + 28(2) = 64
Hence, the 29th term of the AP is 64.