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Question

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

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Solution

Given that,

a3 = 12

a50 = 106

We know that,

an = a + (n − 1) d

a3 = a + (3 − 1) d

12 = a + 2d (I)

Similarly, a50 = a + (50 − 1) d

106 = a + 49d (II)

On subtracting (I) from (II), we obtain

94 = 47d

d = 2

From equation (I), we obtain

12 = a + 2 (2)

a = 12 − 4 = 8

a29 = a + (29 − 1) d

a29 = 8 + (28)2

a29 = 8 + 56 = 64

Therefore, 29th term is 64.


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