let an be an arithmetic sequence. if a4=27 and a9=67, what is a1?
an = a1 + (n-1)d
a4 = a1 + (4-1)d
27 = a1 + 3d
a9 = a1 + (9-1)d
67 = a1 + 8d
So we solve this system of equations:
27 = a1 + 3d
67 = a1 + 8d
To eliminate d, multiply the first
equation through by -8, and multiply
the second equation through by 3
-216 = -8a1 - 24d
201 = 3a1 + 24d
-------------------
-15 = -5a1
3 = a1
--------------------
To check we find d by substituting
3 for a1 in
27 = a1 + 3d
27 = 3 + 3d
24 = 3d
8 = d
So the sequence is found by starting with
3 and adding 8 each time to get the next
term:
3, 11, 19, 27, 35, 43, 51, 59, 67
Edwin