What’s the 33th term of this sequence 19, 15, 11, 7

To solve this problem, we first have to identify if this is an arithmetic sequence or not which in this case it is since the sequence has a common difference which is the number repeatedly subtracted to reach the next term.

One way to solve this problem is to start with 19 which is your first term and keep on subtracting 4 until you reach the 33 term.s However, if you're taking a test, that's probably not the best use of your time.

However, we can also use our explicit formula which is shown below.

For , we will substitute 33 in for n since we are solving for what the 33rd term is in this arithmetic sequence.

Then, is the first term in your sequence which is 19.

Inside the parentheses, n will also be 33.

Lastly, d is your common difference which is -4 since we need to subtract 4 to reach the next term in this sequence.

now we have all the information we need but we still need to simplify so here's where the math comes in.

So we have .

Make sure to apply order of operations because this is where many students make mistakes. So, simplify inside the parentheses first.

33 - 1 is going to be 32 so we have  .

Now multiply before we add so (32)(-4) is -128.

So we have which is -109.

So the 33rd term of this sequence is -109.

3 is greater then both of those numbers