t1 + d = 9/2 t5 = t1 + 4d = 6 Now we have two simultaneous equations: t1 + 4d = 6 ............(1) t1 + d = 9/2 ...........(2) Subtracting (2) from (1) gives: 3d = 3/2 So the common difference d = 1/2 and by substitution in (1) we find t1 = 4 Therefore the 3rd term = 4 + (2 * 1/2) = 5
