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

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#### Solution

Given that,

a_{3} = 12

a_{50} = 106

We know that,

a_{n} = a + (n − 1) d

a_{3} = a + (3 − 1) d

12 = a + 2d (I)

Similarly, a_{50 }= 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

a_{29} = a + (29 − 1) d

a_{29} = 8 + (28)2

a_{29} = 8 + 56 = 64

Therefore, 29^{th} term is 64.

Concept: nth Term of an AP

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