Arithmetic Sequence 4th term

[sabanation12]

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!

 

[sabanation12]

please help me

 

[hyurnat4]

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.

 

[gsal]

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
.
.
.

 

[sabanation12]

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:

a_n = a₁ + (n-1) * d

Plugging in numbers:

a_n = 12 + (4-1) * 18

so the fourth number is 66

Thanks for your help anyways :)