### Video Transcript

Find the arithmetic sequence given the sum of the fifth and 10th terms is negative 297, and the seventh term is four times greater than the fourth term.

First, we need to think about what we know about an arithmetic sequence. If we let the first term be π sub one, in an arithmetic sequence, the second term will be the first term plus the common difference, π, and the third term will be the second term plus the common difference, π. This pattern continues all the way to π sub π. And when we say π sub π, we just mean any term π in the sequence.

Mathematically, if we want to find any term in the sequence, we need to know two things: the first term and the common difference, π. We havenβt been given the first term or the common difference, so letβs consider what we do have to start with. We know that the sum of the fifth and 10th terms is negative 297. Weβll write that as π sub five plus π sub 10 equals negative 297. We also know that the seventh term is four times greater than the fourth term. And so, we can say that π sub seven equals four times π sub four.

In these two equations, though, we have four different variables. And since we only have two equations, we canβt solve for four different variables. We need to try to set these equations in different terms. In fact, we want to rewrite these equations in terms of the first term and the common difference. For example, we know that π sub five is equal to π sub one plus four times π. Since weβre dealing with the fifth term, we multiply our common difference by four.

π sub 10 is then π sub one plus nine π. π sub seven is equal to π sub one plus six π. And π sub four is equal to π sub one plus three π. And now, we want to substitute these new values in place of our four variables. In our first equation, weβll have π sub one plus four π plus π sub one plus nine π equals negative 297. Here, we can combine like terms. Two π sub one plus 13π equals negative 297. And in our second equation, we have π sub one plus six π is equal to four times π sub one plus three π. We need to distribute the four, which gives us four π sub one plus 12 π.

With this equation, we might want to get the π sub ones on one side and the π-variable on the other side. So, we subtract six π from both sides. Then, we have π sub one equals four π sub one plus six π. From there, we can subtract four π sub one from both sides of the equation, which tells us that negative three π sub one equals six π. And then, we divide both sides of the equation by negative three. And we see that π sub one, the first term, is equal to negative two π.

What we want to do now is substitute negative two π in for π sub one in our first equation so that we have two times negative two π plus 13π equals negative 297, which is negative four π plus 13π equals negative 297. Again, when we combine like terms, weβll have nine π. To solve for π, we need to divide both sides of the equation by nine. And this tells us that π, our common difference, is negative 33.

Remember, we said we need to know two things to find every term in the sequence. We now know the common difference, and we can use that common difference to find the first term. We know the first term is equal to negative two times the common difference. And that means the first term is negative two times negative 33. Our first term here is 66.

Now that we know these two pieces of information, weβre ready to list out the sequence. Its first term is 66 and the second term will be the first term plus the common difference. 66 plus negative 33 equals positive 33. The third term will be the second term plus the common difference. 33 plus negative 33 equals zero. And the sequence would continue in this way.