Question

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

Answer 1

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Answer 2

According to the Question

$\bf\huge n = 50 , \:a_{3} = 12 \:and\: a_{50} = 106$

$\bf\huge a_{3} = 12$

12 = a + 2d …… (1)

$\bf\huge a_{50} = 106$

106 = a + 49d …… (2)

Now Subtract (1) from (2) we get

94 = 47d

$\bf\huge\frac{94}{47} = d$

d = 2

Put value in (1)

12 = a + 2(2)

12 - 4 = a

a = 8

$\bf\huge a_{29} = a + 28d$

8 + 28(2) = 64

Hence 29th term is 64