Question

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.

Answer 1

Let a , d are first term and common difference of an A.P
nth term = Last term = a + ( n - 1 )d

an = a + ( n - 1 )d
Now ,

It is given that ,

Third term = 12

a + 2d = 12 ------( 1 )

Last term = 106

a + 49d = 106 ---( 2 )

Subtract ( 1 ) from ( 2 ) , we get

47d = 94

d = 2

Substitute d value in equation ( 1 ) ,

We get

a + 2 × 2 = 12

a = 12 - 4

a = 8

29th term = a + 28d

a29 = 8 + 28 × 2

= 8 + 56

= 64