Question

If the 3rd term and the last term of an A. P are 12 and 106. Find the 29th term of the no. of term is 50.

Answer 1

hii

here is ur solution


$a3 = a + (3 - 1)d \\ 12 = a + 49d.......(1)$



$a50 = a + (50 - 1)d \\ 106 = a + 49d......(2)$




subtract (1) from the (2)


$106 = a + 49d \: - 12 = a + 2d = - 94= - 47d \: \\ d = 2$
from eq (1)

$12 = a + (2 \times 2) \\ a = 12 - 4 = 8$

for 29th term



$a29 = 2 + 28 \times 2 \\ a29 = 64$

hope it helps you