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

An AP consists of 50 terms and the \(50^{th}\) term is equal to 106 and \(a_3 = 12\)

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(\Rightarrow a_{50}\)= a + (50 - 1)d, and
\(a_3\)= a + (3 - 1)d
\(\Rightarrow \) 106 = a + 49d and,
12 = a + 2d

These are equations consisting of two variables.

Using equation 106 = a + 49d,
we get a = 106 - 49d

Putting value of a in the equation 12 = a + 2d,

\(\Rightarrow \) 12 = 106 - 49d + 2d
\(\Rightarrow \) 47d = 94
\(\Rightarrow \) d = 2

Putting value of d in the equation,
a = 106 - 49d,

a = 106 – 49 (2)
= 106 – 98 = 8

Therefore, First term = a = 8 and
Common difference = d = 2

To find \(29^{th}\) term,
we use formula \(a_n = a + (n - 1)d\) which is used to find nth term of arithmetic progression,

\(a_{29}\)
= 8 + (29 - 1) 2
= 8 + 56 = 64

Therefore, 29th term of AP is equal to 64.