Question

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

NCERT Class X
Mathematics - Mathematics

Chapter _ARITHMETIC PROGRESSIONS

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.

: )

Answer 2

Heya !!!




Third term = 12



A + 2D = 12 ---------(1)



It is given that the AP consists 50 term. So the last term of AP should be 50.



50 term = 106



A + 49D = 106 --------(2)

From equation (1) we get,




A + 2D = 12



A = 12-2D ----------(3)



Putting the Value of A in equation (3) we get,



A + 49D = 106




12-2D + 49D = 106





47D = 106-12




47D = 94



D = 94/47




D = 2


Putting the Value of D in equation (3)


A => 12-2D = 12-2 × 2



A = 12-4 = 8




Therefore,


★ 29th term of AP




=> A +28D



=> 8 + 28 × 2




=> 8 + 56




=> 64.







HOPE IT WILL HELP YOU....... (+_+)