Find the 31^st term of an AP whose 11^thterm is 38 and the 16^th term is 73

Given

11^thterm = 38

⇒ a + (11-1)d = 38 → 1

16^th term = 73

⇒ a + (16-1)d = 73 → 2

By subtracting the both equations we will get ‘d’

(a +10d) – (a+15d) = 38 -73

-5d = -35

d = 7

By substituting “d” in equation 1

a +10d = 38

a + 10×7 = 38

a = -32

a₃₁ = a + (31-1) d

= -32 + 30×7

= 178