Question

An AP consists of 50 terms of which 3rd term is 12 and last term is 106.Find the 29th term(Question is from arithmetic progressions class 10)

Answer 1

Hi ,

Let a , d are first term and common

difference of an A.P

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nth term = Last term = a + ( n - 1 )d

an = a + ( n - 1 )d

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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

I hope this helps you.

: )