The 3^rd and 9^th term of an arithmetic progression are -8 and 10 respectively. What is the 16^th term?

Solution (By Examveda Team)

Let the first term of an AP = a and the common difference = d
3^th term of AP = A₃ = a + 2d = -8 ......(1)
9^th term = A₉ = a + 8d = 10 ...... (2)
Subtracting equation (1) from (2), we get :
⇒ 8d - 2d = 10 - (-8)
⇒ 6d =18
⇒ d = $$\frac{{18}}{6}$$ = 3
Substituting it in equation (2),
⇒ a = 10 - 8(3)
      = 10 - 24
      = -14
∴ 16^th term = A₁₆ = a + 15d
= -14 + 15(3)
= -14 + 45
= 31