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

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

Given,

⇒ a₃ = a + (3 - 1)d

⇒ a + 2d = 12 ……..(i)

⇒ a₅₀ = a + (50 - 1)d

⇒ a + 49d = 106 ……..(ii)

Subtracting (i) from (ii) we get,

⇒ a + 49d - (a + 2d) = 106 - 12

⇒ 49d - 2d = 94

⇒ 47d = 94

⇒ d = 2.

Substituting value of d in (i) we get,

⇒ a + 2(2) = 12

⇒ a + 4 = 12

⇒ a = 8.

a₂₉ = a + (29 - 1)d

a + 28d = 8 + 28(2) = 8 + 56 = 64.

Hence, 29^th term = 64.