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

Let first term of A.P. be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

Last term of A.P. = 106 and there are 50 terms in the A.P.

⇒ a₅₀ = 106

⇒ a + (50 - 1)d = 106

⇒ a + 49d = 106 …….(1)

Given,

3rd term = 12

⇒ a₃ = 12

⇒ a + (3 - 1)d = 12

⇒ a + 2d = 12 …….(2)

Subtracting equation (2) from (1), we get :

⇒ a + 49d - (a + 2d) = 106 - 12

⇒ a - a + 49d - 2d = 94

⇒ 47d = 94

⇒ d = $\dfrac{94}{47}$ = 2.

Substituting value of d in equation (2), we get :

⇒ a + 2 × 2 = 12

⇒ a + 4 = 12

⇒ a = 12 - 4

⇒ a = 8.

29th term = a₂₉

= a + (29 - 1)d

= 8 + 28 × 2

= 8 + 56

= 64.

Hence, 29th term = 64.