Find the 3rd term of an arithmetic sequence with t2 = 9/2 and t5 = 6?

5

4

3

2

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