Q. 84.3( 232 Votes )

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

Answer :

Given, a_{3} = 12 and a_{50} = 106

a

_{n}= a + (n - 1)d

where, a

_{n}= nth term of the AP

a = first term of AP

n = number of terms of AP

d = common difference of AP

Applying the formula for 3rd and 50th term we get,

a_{3} = a + 2d = 12

a_{50} = a + 49d = 106

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

a + 49d – a – 2d = 106 – 12

47d = 94

d = 2

Substituting the value of d in 3rd term, we get;

a + 2 x 2 = 12

Or, a + 4 = 12

Or, a = 8

Now, 29th term can be calculated as follows:

a_{29} = a + 28d

= 8 + 28 x 2

= 8 + 56 = 64

a_{29}= 64

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