Question

The 3rd term of an ap is 8 the 9th term of an ap exceeds the third term by 2 find the sum of first 19 terms

Answer 1

Answer:

589/3

Step-by-step explanation:

Given t₃ = 8 & t₉ = t₃+2 = 8+2 =10 then find S₁₉

t₃ = a+2d = 8 -------- (1)

t₉ = a+8d = 10 --------(2)

(1) -(2) =>

=> a+2d - (a+8d) = 8-10

=> 2d-8d = -2

=> -6d = -2

=> d=1/3

substitute d=1/3 in (1), we get

a+2(1/3) = 8

a = 8-(2/3)

  = (24-2)/3

a  = 22/3

therefore , S₁₉ = (19/2) [2(22/3)+18(1/3)]   by formula Sₙ= n/2 [2a+(n-1)d]

                       = (19/2) [(44/3)+(18/3)]

                       = (19/2)[62/3]

                      = (19*31)/3

                      = 589/3