Video: Finding the Arithmetic Sequence under a Given Condition

Find the arithmetic sequence in which π‘Žβ‚‚ + π‘Žβ‚„ = βˆ’28 and π‘Žβ‚ƒ Γ— π‘Žβ‚… = 140.

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Video Transcript

Find the arithmetic sequence in which π‘Ž sub two plus π‘Ž sub four is equal to negative 28 and π‘Ž sub three times π‘Ž sub five equals 140.

Let’s think about how these kinds of sequences work. If we let π‘Ž sub one be the first term in our sequence, π‘Ž sub two be the second, and so on, to get from term one to term two, we have a difference of 𝑑. And so we say, in an arithmetic sequence, the terms have a common difference of 𝑑. And that means we can find term 𝑛 by taking the first term, π‘Ž sub one, and adding 𝑛 minus one times 𝑑. We can think about it like this. π‘Ž sub four will be equal to π‘Ž sub one plus three 𝑑.

Now that we know that, we can consider the information we were given. We’re told that π‘Ž sub two plus π‘Ž sub four equals negative 28. The second term plus the fourth term equals negative 28. We also know that π‘Ž sub three times π‘Ž sub five equals 140. The third term times the fifth term equals 140. The first thing we can do is write all of our terms in relationship to π‘Ž sub one and 𝑑. Here’s what I mean by that. We can rewrite π‘Ž sub two as π‘Ž sub one plus the common difference 𝑑. And that means π‘Ž sub three is equal to the first term plus two times the common difference. The fourth term equals the first term plus three times the common difference. And the fifth term equals the first term plus four times the common difference.

Looking at our first equation, we want to substitute this relationship 𝑛 for the second and fourth terms. In place of the second term π‘Ž sub two, we’ll have π‘Ž sub one plus 𝑑. And in place of the fourth term, we’ll have π‘Ž sub one plus three 𝑑, all equal to negative 28. We can combine like terms. We have π‘Ž sub one plus π‘Ž sub one, which gives us two π‘Ž sub one. And then, we can combine 𝑑 plus three 𝑑 to give us four 𝑑. Now we can say that two π‘Ž sub one plus four 𝑑 equals negative 28. At this point, we recognize that we have three even terms. And so we can divide the entire equation by two, which will give us π‘Ž sub one plus two 𝑑 is equal to negative 14.

At this point, you might not recognize anything else that we can do. So we move on to our second equation. And we substitute π‘Ž sub one plus two 𝑑 in for π‘Ž sub three. And we substitute π‘Ž sub one plus four 𝑑 in for the fifth term. While you’re doing that, hopefully, you recognize this term π‘Ž sub one plus two 𝑑. We’ve already found out what π‘Ž sub one plus two 𝑑 is equal to, negative 14. We can substitute negative 14 in for π‘Ž sub one plus two 𝑑. And if we look closely, that’s because π‘Ž sub one plus two 𝑑 is equal to the third term in the sequence. And so we can plug that value in for the third term of our sequence.

Now, what should we do? We could go back to our first equation, π‘Ž sub one plus two 𝑑 equals negative 14, and subtract two 𝑑 from both sides, which tells us that π‘Ž sub one is equal to negative 14 minus two 𝑑. And then, we could plug in negative 14 minus two 𝑑 in for π‘Ž sub one. If we do this, we combine like terms. Negative two 𝑑 plus four 𝑑 equals two 𝑑. So we have negative 14 times negative 14 plus two 𝑑 equals 140. Then, we divide both sides of the equation by negative 14. And we see that negative 14 plus two 𝑑 equals negative 10. So we add 14 to both sides. And we find that two 𝑑 equals four. So we divide both sides by two. And we find that 𝑑, our common difference, equals two.

Let’s go back to this π‘Ž sub three times π‘Ž sub five for a minute, which equals 140. Once we found out that the third term was negative 14, we can very quickly and easily find the fifth term by dividing both sides of this equation by negative 14. π‘Ž sub five equals negative 10. Now that we know our fifth term is negative 10, we’ll plug that in. To solve the rest of our terms, we can just use our common difference. We know that each term we’re adding two. So our fourth term is going to be negative 14 plus two, which is negative 12. To find our second term, we need to work backwards. We need to say what plus two equals negative 14. Another way to say that is 14 minus two, which is negative 16. And again, for our first term, we’re asking what plus two equals negative 16 or what is negative 16 minus two. And that’s negative 18.

We found the first five terms in this sequence. But at this point, it’s still probably worth checking that the two statements we started with are true. We were told that the second and fourth terms combined should equal negative 28. The second term is negative 16. And the fourth term is negative 12. Negative 16 plus negative 12 does equal negative 28. We were told that the third term times the fifth term equals 140. That’s negative 14 times negative 10, which does equal 140.

We note a sequence like this in parentheses, where we list the first, second, and third term: negative 18, negative 16, negative 14 continuing.

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