Question

An arithmetic progression has50 terms ,the third term is12 and the last term is 106 determining the 29th term

Answer 1

Answer:

let the 1st term is a and common difference is b

a+2b=12

a =12-2b

again,

a+49b=106

12-2b+49b=106

47b=106-12=94

b=94/47=2

a =12-2.2=12-4=8

so 29 th term is

a+28b

8+28.2=8+56=64