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
