Question

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

Answer 1

Given, a3 = 12 and a50 = 106

a3 = a + 2d = 12

a50 = a + 49d = 106

Subtracting 3rd term from 50th term, we get;

a + 49d – a – 2d = 106 – 12

Or, 47d = 94

Or, d = 2

Substituting the value of d in 12th term, we get;

a + 2 x 2 = 12

Or, a + 4 = 12

Or, a = 8

Now, 29th term can be calculated as follows:

a29 = a + 28d

= 8 + 28 x 2

= 8 + 56 = 64

Answer 2

Given,

In A.P.,

n= 50

a3= 12,

a50= 106

a29=?

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

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

From eq (1) and (2),

47d= 94

d= 2

Putting value of d in eq. (1), we get

a+2(2)= 12

a+4= 12

a= 8

Now,

a29= a+28d

a29= 8+28(2)

a29= 8+56

a29= 64.

Answer is 64.

Hope it helps ✌✌❤