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Lesson Video: Applications of Arithmetic Sequences Mathematics • 9th Grade

In this video, we will learn how to solve real-world applications of arithmetic sequences, where we will find the common difference, nth term explicit formula, and order and value of a specific sequence term.

15:48

Video Transcript

In this video, we will learn how to solve problems involving real-world applications of arithmetic sequences. We’ll see how to find the common difference, the explicit formula for the 𝑛th term, and the order or value of a specific term in an arithmetic sequence.

Let’s begin by recalling what an arithmetic sequence is. Well, it’s an ordered list of terms in which the difference between consecutive terms is constant. For example, the four times table, four, 8, 12, 16, and so on, is an example of an arithmetic sequence because the difference between each pair of consecutive terms is four. We call this difference the common difference, and we represent it using the letter 𝑑. We often use the letter π‘Ž to represent the first term in the sequence and we have a formula for calculating the general or 𝑛th term π‘Ž sub 𝑛, the 𝑛th term, is equal to π‘Ž plus 𝑛 minus one multiplied by 𝑑.

This tells us that to calculate any subsequent term in the sequence, we can take the first term and add on 𝑛 minus one lots of the common difference 𝑑, which makes sense if we think about it. To find the second term, we need to add the common difference on once. To find the third term, we have to add the common difference on twice, so we’re always adding on one less lot of the common difference than the term number. We also have a formula for calculating the sum of the first 𝑛 terms in an arithmetic sequence. 𝑆 sub 𝑛 is equal to 𝑛 over two multiplied by two π‘Ž plus 𝑛 minus one 𝑑 where, again, π‘Ž represents the first term in the sequence and 𝑑 represents the common difference.

So, those are the basics of arithmetic sequences which we should already be familiar with. Let’s now look at how we can apply these results to some real-world problems. In our first example, we’ll see how we can find a specific term in an arithmetic sequence which is presented to us as a word problem.

Mason’s exercise plan lasts for six minutes on the first day and increases by four minutes each day. For how long will Mason exercise on the 18th day?

We can see that Mason increases his exercise plan by the same amount of four minutes each day. This means that the times Mason spends exercising daily form an arithmetic sequence with a common difference of four. We’re also told that Mason spends six minutes exercising on the first day of his plan, which means that the first term of this arithmetic sequence π‘Ž is equal to six. We therefore have all the information we need to write down as many terms of the sequence as we want or write down the rule for the 𝑛th term.

The first term in this sequence is six. The second term is four more than this, so it’s 10. The third term is four more than this, so it’s 14. We could continue in this way, but it’s not very efficient if we need to get all the way to the 18th term in this sequence. Instead, we can use the formula for the 𝑛th term: π‘Ž sub 𝑛 is equal to π‘Ž plus 𝑛 minus one multiplied by 𝑑. Substituting six for π‘Ž, the first term, and four for 𝑑, the common difference, we have π‘Ž sub 𝑛 is equal to six plus four multiplied by 𝑛 minus one.

We could simplify this algebraically, or to find the 18th term, we could go straight to substituting 𝑛 equals 18. π‘Ž sub 18 is equal to six plus four multiplied by 18 minus one. We have six plus four multiplied by 17. Four multiplied by 17 is 68. And adding six gives 74. Remember that the terms in the sequence are times in minutes. So we found that the 18th term of this sequence or the time Mason spends exercising on the 18th day is 74 minutes.

So in this example, we’ve seen how to calculate a particular term in an arithmetic sequence. In our next example, we’ll see how we can work out the order or term number of a particular term, again from a worded problem.

Olivia is training for a 10-kilometer race. On each training day, she runs 0.5 kilometers more than the previous day. If she completes four kilometers on her fourth day, on what day will she complete 10 kilometers?

Let’s look carefully at the information we’ve been given. We’re told that on each training day, Olivia runs 0.5 or half a kilometer more than the previous day. This means that the distances Olivia runs each day form an arithmetic sequence with a common difference 𝑑 of 0.5. We don’t know how far Olivia ran on the first day. But we do know that she runs four kilometers on the fourth day. We can therefore use the formula for the general term of an arithmetic sequence π‘Ž sub 𝑛 equals π‘Ž plus 𝑛 minus one 𝑑 to form an equation. We have four is equal to π‘Ž plus 0.5 multiplied by four minus one. That simplifies to four is equal to π‘Ž plus 1.5. And we can solve this equation for π‘Ž by subtracting 1.5 from each side.

Doing so, we have π‘Ž is equal to 2.5. So we now know that Olivia ran 2.5 kilometers on the first day of her training. What we’re asked, though, is on what day will she complete 10 kilometers? So which term in the sequence or what value of 𝑛 gives a term equal to 10? We can therefore substitute π‘Ž equals 2.5, 𝑑 equals 0.5, and π‘Ž 𝑛 equals 10 to give us an equation we can solve to find the value of 𝑛. Distributing the parentheses, we have 2.5 plus 0.5𝑛 minus 0.5 equals 10. And then the left-hand side simplifies to two plus 0.5𝑛 is equal to 10. We can subtract two from each side to give 0.5𝑛 is equal to eight and then multiply each side of our equation by two to give 𝑛 is equal to 16. So the term number or the order of the term which is equal to 10 is 16. And so we know that Olivia will complete 10 kilometers on the 16th day of her training plan.

Of course, the other way to answer this question once we’d calculated the value of π‘Ž would’ve been to list out all the terms of the sequence by adding 0.5 each time: 2.5, three, 3.5, four. But it would take quite a long time to get to the 16th term, so it’s more efficient to use the first method. In either case so, our answer to the problem is 16.

In our next example, we’ll see how to calculate a specific term in an arithmetic sequence when we’ve been given some information about two of the other terms.

Amelia’s annual salary increases by the same quantity every year. In her fourth year at her job, she earned 24,000 dollars. In her 10th year, she earned 36,000 dollars. How much will she earn in her 20th year?

If Amelia’s annual salary increases by the same amount every year, then her annual salaries form an arithmetic sequence. We can therefore express the general term in this sequence using the rule π‘Ž sub 𝑛 is equal to π‘Ž plus 𝑛 minus one 𝑑 where π‘Ž represents the first term in the sequence, so Amelia’s salary in her first year, and 𝑑 represents the common difference. That’s the annual increase. We don’t know either of these values, but instead we’ve been given some information about Amelia’s salary in the fourth and 10th years. We can use this information to form some equations. In the fourth year, she earned 24,000 dollars. So we have the equation 24,000 equals π‘Ž plus four minus one 𝑑 or more simply π‘Ž plus three 𝑑.

We’re also told that in the 10th year, she earned 36,000 dollars. So we also have the equation 36,000 equals π‘Ž plus 10 minus one 𝑑 or π‘Ž plus 9𝑑. What we now have is a pair of linear simultaneous equations with two unknowns, π‘Ž and 𝑑. And so we can solve these two equations simultaneously. By subtracting equation one from equation two, the π‘Ž-terms will cancel, and we’re left with 12,000 is equal to 6𝑑. Dividing through by six, and we found the common difference for this sequence. 𝑑 is equal to 2,000. So that’s Amelia’s annual pay increase.

To find the value of π‘Ž, we can sub 𝑑 equals 2,000 into either of the two equations. I’ve chosen equation one, giving 24,000 equals π‘Ž plus three multiplied by 2,000. Subtracting 6,000, that’s three multiplied by 2,000, from each side, and we have the value of π‘Ž. π‘Ž is equal to 18,000. So that was Amelia’s salary in the first year of her job. What we’re asked to find, though, is what Amelia will earn in her 20th year. So we need to find the 20th term of this sequence. We can do this by substituting the values of π‘Ž and 𝑑 and the value 𝑛 equals 20 into our general term formula. We have π‘Ž sub 20 is equal to 18,000 plus 19. That’s 20 minus one multiplied by 2,000. That’s 18,000 plus 38,000, which is 56,000. We found then in the 20th year of her job, Amelia will earn 56,000 dollars assuming she continues to get the same pay increase of 2,000 dollars every year.

So we’ve seen some examples of how we can calculate a specific term or the order of a term in an arithmetic sequence. In our next example, we’ll practice finding the sum of the terms in an arithmetic sequence which has been presented as a word problem.

A runner is preparing himself for a long-distance race. He covered six kilometers on the first day and then increased the distance by 0.5 kilometers every day. Find the total distance he covered in 14 days.

As the distance run increases by the same amount every day, these distances form an arithmetic sequence. The common difference for this sequence is 0.5, and the first term π‘Ž is the distance run on the first day. That’s six kilometers. To find the total distance covered in 14 days, we need to find the sum of the first 14 terms in this sequence. We recall then that the sum of the first 𝑛 terms in an arithmetic sequence can be found using the formula 𝑆 sub 𝑛 is equal to 𝑛 over two multiplied by two π‘Ž plus 𝑛 minus one 𝑑. We can therefore substitute 14 for 𝑛, six for π‘Ž, and 0.5 for 𝑑, giving 𝑆 sub 14 is equal to 14 over two multiplied by two times six plus 0.5 multiplied by 14 minus one.

That simplifies to seven multiplied by 12 plus 0.5 multiplied by 13. We keep going inside the parentheses. We have 12 plus 6.5, which is 18.5, and then multiplying by seven gives 129.5. Remember, this is a distance and the units are kilometers. So by applying the formula for the sum of the first 𝑛 terms in an arithmetic sequence, we found the total distance covered by this runner in 14 days is 129.5 kilometers.

In our final example, we’ll see how we can find the 𝑛th term rule for an arithmetic sequence which once again will be presented as a word problem.

Emma started a workout plan to improve her fitness. She exercised for 14 minutes on the first day and increased the duration of her exercise plan by six minutes each subsequent day. Find, in terms of 𝑛, the 𝑛th term of the sequence which represents the number of minutes that Emma spends exercising each day. Assume that 𝑛 equals one is the first day of Emma’s plan.

We’re told in this problem that Emma increased her exercise by the same amount every day, which means that the time spent exercising form an arithmetic sequence with a common difference of six. We’re also told that Emma exercised for 14 minutes on the first day of her plan, which means the first term in the sequence is 14. We are asked to find in terms of 𝑛 the 𝑛th term of this sequence, so we need to recall the general formula for the 𝑛th term of an arithmetic sequence. It’s this: π‘Ž sub 𝑛, the 𝑛th term, is equal to π‘Ž plus 𝑛 minus one 𝑑, where π‘Ž represents the first term and 𝑑 represents the common difference.

We can therefore substitute the values of π‘Ž and 𝑑, which we were given in the question to find our general term. It’s π‘Ž sub 𝑛 is equal to 14 plus six multiplied by 𝑛 minus one. Now it’s usual to go on and simplify algebraically. So we’ll distribute the parentheses. We have 14 plus six 𝑛 minus six, which then simplifies to six 𝑛 plus eight. And it is usual to give the general term of an arithmetic sequence in this form, some multiple of 𝑛 plus a constant. Notice as well that that common difference of six is the coefficient of 𝑛 in our general term and that will always be the case for an arithmetic sequence. We found the 𝑛th term of this sequence. It’s six 𝑛 plus eight. And by substituting any value of 𝑛, we can calculate any term in this sequence.

Let’s now review some of the key points that we’ve covered in this video. Firstly, we reminded ourselves that in an arithmetic sequence, the difference between consecutive terms is constant and we call this the common difference and represent it using the letter 𝑑. The first term of an arithmetic sequence is usually denoted by the letter π‘Ž although it can also be denoted as π‘Ž subscript one. We then denote subsequent terms in the same way: π‘Ž sub two, π‘Ž sub three, and so on. We have a formula for calculating the general or 𝑛th term in an arithmetic sequence using the first term, the common difference, and the term number. π‘Ž sub 𝑛 is equal to π‘Ž plus 𝑛 minus one multiplied by 𝑑.

We also have a formula for calculating the sum of the first 𝑛 terms in an arithmetic sequence: 𝑆 sub 𝑛 is equal to 𝑛 over two multiplied by two π‘Ž plus 𝑛 minus one 𝑑. In this video, we saw specifically how we can apply these results to word problems. We must ensure that we read the question carefully, identify key information, and then form any equations. We can then solve these equations to find the general term, a specific term, the order of a term, or the sum of the first 𝑛 terms in any arithmetic sequence.

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