# Arithmetic Sequence 4th term

## Main Question or Discussion Point

Our 8th grade math counts team met today and I didnt know how to do this problem:

The first three terms of an arithmetic sequence are p, 2p+6, and 5p-12. What is the 4th term of this sequence?

Please explain how to do this.

Arigato!

Its not an arithmetic progression... and don't bump. In an AP the difference between consecutive terms is constant (i.e. t1-t0=t2-t1 and so forth.) In this case $$(2p+6)-p=5p-12-(2p+6)$$
$$p+6=3p-18$$
$$2p=24$$
$$p=12$$
Only true when p=12. For all other cases it is not an AP.

so, is that the solution? p=12 and a difference of 18 between two consecutive numbers?

2p-12 <- simplifies to p, of course, if you know p=12
2p+6
5p-12
5p+6
8p-12
8p+6
.
.
.

Its not an arithmetic progression... and don't bump. In an AP the difference between consecutive terms is constant (i.e. t1-t0=t2-t1 and so forth.) In this case $$(2p+6)-p=5p-12-(2p+6)$$
$$p+6=3p-18$$
$$2p=24$$
$$p=12$$
Only true when p=12. For all other cases it is not an AP.
No I believe you are wrong, it IS an arithmetic sequence. Let me explain...

P does equal 12, and the difference between them is 18, so:

an = a1 + (n-1) * d

Plugging in numbers:

an = 12 + (4-1) * 18

so the fourth number is 66

Thanks for your help anyways :)