Question

An ap consists of 50 terms in its third term is 12 and last term is 106 find the 29th term of the ap

Answer 1

Answer:

n=50

a3=12

a50=106

a29=?

$a3 = a + (n - 1)d \\ 12 = a + (3 - 1)d \\ 12 = a + 3d - d \\ 12 = a + 2d \\ 12 - 2d = a..........1 \\ \\a50 = a +( n - 1)d \\ 106 = a + (50 - 1)d \\ 106 = a + 49d \\ 106 - 49d = a........2 \\ put \: 2in1 \\ 12 - 2d = 106 - 49d \\ 12 - 106 = - 49 d+ 2d \\ 94 = 47d \\ d = \frac{94}{47} \\ d = 2 \\ put \: d \: in \: 1 \\ 12 - 2d = a \\ 12 - 2(2) = a \\ 12 - 4 =a \\ 8 = a \\ a29 = a + (n - 1)d \\ = 8 + 28(2) \\ = 8 + 56 \\ = 64$