Video: Finding the Arithmetic Sequence under a Certain Condition

Find the arithmetic sequence given the fourth term is 42 and the sum of the fifth and ninth terms is 12. Then find the order of the term whose value is βˆ’330 in this sequence.

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Video Transcript

Find the arithmetic sequence given the fourth term is 42 and the sum of the fifth and ninth terms is 12. Then find the order of the term whose value is negative 330 in this sequence.

First, let’s write what we know. The fourth term in the sequence is 42. So we write that as π‘Ž sub four equals 42. The sum of the fifth and ninth terms is 12. We can write that out as π‘Ž sub five plus π‘Ž sub nine equals 12. And this is where we need to think about what we know about arithmetic sequences. π‘Ž sub 𝑛, so any term in the sequence, is equal to π‘Ž sub one, the first term, plus 𝑛 minus one times 𝑑. This means we can find any term in the sequence if we know the first term, π‘Ž sub one, and 𝑑, the common difference.

In order to solve this problem, we need to set up a system of equations. Let’s start with π‘Ž sub four. We know that π‘Ž sub four will be equal to the first term plus three times 𝑑. In the same way, we can say the fifth term will be equal to π‘Ž sub one plus four 𝑑 and the ninth term will be equal to π‘Ž sub one plus eight 𝑑. And we want to take this information and plug it in to the equations we started with.

We’ll have an equation that says π‘Ž sub one plus three 𝑑 equals 42 and another that says π‘Ž sub one plus four 𝑑 plus π‘Ž sub one plus eight 𝑑 equals 12. Here, we can combine like terms so that we have two π‘Ž sub one plus 12𝑑 equals 12. We notice that we have coefficients that are all even. To simplify this equation, we can divide through by two and have the equation π‘Ž sub one plus six 𝑑 equals six.

What we want to do now is take our first equation and plug it in to our second equation. If we solve our first equation for π‘Ž sub one, we would subtract three 𝑑 from both sides and say that π‘Ž sub one equals 42 minus three 𝑑. And if π‘Ž sub one equals 42 minus three 𝑑, we can plug in 42 minus three 𝑑 in place of π‘Ž sub one in our second equation, which means we’ll have 42 minus three 𝑑 plus 60 equals six. If we do a bit more of combining like terms, negative three 𝑑 plus six 𝑑 equals three 𝑑.

Again, we’re trying to solve for 𝑑. So we subtract 42 from both sides, and we get three 𝑑 equals negative 36. So we divide through by three, and we find that 𝑑 equals negative 12. Remember, at the beginning, I said that we needed the common difference and the first term in order to be able to find any term in the sequence. We have the common difference, so we need to solve for our first term. And we can do that by taking negative 12 and plugging it in for 𝑑 in our first equation so that we have π‘Ž sub one equals 42 minus three times negative 12.

42 plus 36 equals 78. This means the first term of this sequence is 78 and the common difference is negative 12. The first step in this problem was to find the arithmetic sequence. We know its first term is 78. And since it has a common difference of negative 12, the second term will be 12 less than the first term. 78 minus 12 is 66. The third term will be 12 less than the second term. 66 minus 12 is 54. We generally list a sequence by its first three terms and then indicate that it continues with the dots at the end. Here’s our sequence.

But for part two, we need to know what term is equal to negative 330. Negative 330 is going to be equal to π‘Ž sub one plus 𝑛 minus one times 𝑑. We know that 𝑑 equals negative 12 and π‘Ž sub one equals 78. To solve this value, we distribute our negative 12 so that we have 78 minus 12𝑛 plus 12. We can add 78 plus 12 together, combine like terms. We’ll just slide everything up to give ourselves a little bit more room.

78 plus 12 is 90. So we have negative 330 equals 90 minus 12𝑛. We’re trying to solve for 𝑛, so we subtract 90 from both sides of the equation. And we get negative 420 equals negative 12𝑛. We divide both sides of the equation by negative 12, and we get 𝑛 equals 35. This is telling us that when 𝑛 equals 35, the sequence is equal to negative 330. We are saying that the 35th term of this sequence equals negative 330, which makes the second part of our answer π‘Ž sub 35.

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