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

Given AP has 50 terms,

So general term can be written as,

a_n = a + (n – 1) d………..(i)

Now given the AP’s 3^rd term is 12, substituting this in general term, we get

a₃ = a + (3 – 1)d = 12

⇒ a + 2d = 12…………(ii)

It is also given the last term is 106, as we know the AP has 50 terms, so 50^th term is 106, substituting this in general term, we get

a₅₀ = a + (50 – 1)d = 106

⇒ a + 49d = 106…………(iii)

Now subtracting equation (ii) from equation (iii), we get

⇒ d = 2

Now substituting the value of d in equation (ii), we get

⇒ a + 2d = 12

⇒ a + 2(2) = 12

⇒ a + 4 = 12

⇒ a = 12 – 4

⇒ a = 8

So, the given AP has first term, a = 8 and the difference, d = 2, so the 29^th term will have n = 29, substituting these values in equation (i), we get

⇒ a₂₉ = a + (29 – 1)d

⇒ a₂₉ = 8 + (28)2

⇒ a₂₉ = 8 + 56

⇒ a₂₉ = 64

Hence the 29^th term of the given AP is 64.