Lesson Video: Finding the Arithmetic Sequence Mathematics

In this video, we will learn how to find arithmetic sequences given information about their terms and relationships between them.

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Video Transcript

In this video, we will learn how to find arithmetic sequences given information about their terms and the relationships between them. We recall that an arithmetic sequence or arithmetic progression is a sequence of numbers such that the difference between any two consecutive terms in the sequence is constant. We call this constant difference the common difference. If the sequence has a finite number of terms, then we call it a finite arithmetic sequence. Otherwise, we refer to it as just an arithmetic sequence. Let’s consider an example.

If the first term in our arithmetic sequence is 100 and the common difference between any two terms is negative five, then we can find the next term in the sequence. We will call the first term in the sequence π‘Ž sub one and the second term π‘Ž sub two. As the difference between the second and first terms is negative five, π‘Ž sub two minus 100 equals negative five. Adding 100 to both sides of this equation, we see that π‘Ž sub two is equal to negative five plus 100. π‘Ž sub two, the second term in our arithmetic sequence, is therefore equal to 95.

We can continue this pattern to find an infinite number of terms of the sequence. This is equivalent to saying we can find the next term in the sequence by adding the common difference to the previous term. π‘Ž sub two is therefore equal to π‘Ž sub one plus 𝑑. π‘Ž sub three, the third term, is equal to π‘Ž sub two, the second term, plus the common difference 𝑑. And π‘Ž sub four is equal to π‘Ž sub three plus 𝑑. This pattern continues.

If we denote the 𝑛th term in any arithmetic sequence as π‘Ž sub 𝑛, then we can find a formula for the 𝑛th term in this arithmetic sequence. Going back to our expressions for π‘Ž sub two and π‘Ž sub three, since π‘Ž sub two is equal to π‘Ž sub one plus 𝑑, then π‘Ž sub three is equal to π‘Ž sub one plus 𝑑 plus 𝑑. Simplifying this, we see that π‘Ž sub three is equal to π‘Ž sub one plus two 𝑑. Continuing this pattern, π‘Ž sub four is equal to π‘Ž sub one plus three 𝑑. The formula for the 𝑛th term of an arithmetic sequence with common difference 𝑑 and first term π‘Ž sub one is therefore given by π‘Ž sub 𝑛 is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑.

We will now use this formula to determine information about arithmetic sequences by using the terms of the sequence. Let’s begin by looking at some examples of finding an arithmetic sequence given information about its terms.

Find the finite sequence π‘Ž sub 𝑛, given π‘Ž sub one is equal to negative 82, π‘Ž sub 12 is equal to negative 203, and the twelfth-to-last term is negative 115.

We begin by recalling that as our sequence is finite, it will have a fixed number of terms. We also recall that an arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant. This is called the common difference. For any arithmetic sequence, the formula for the 𝑛th term states that π‘Ž sub 𝑛 is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑, where π‘Ž sub 𝑛 is the 𝑛th term, π‘Ž sub one is the first term, and 𝑑 is the common difference.

In this question, we are told that π‘Ž sub one, the first term, is negative 82. The twelfth term, π‘Ž sub 12, is equal to negative 203. Using the general formula, we see that π‘Ž sub 12 is equal to π‘Ž sub one plus 12 minus one multiplied by 𝑑. Substituting in the values of π‘Ž sub 12 and π‘Ž sub one gives us the equation negative 203 is equal to negative 82 plus 11𝑑. Adding 82 to both sides of this equation, we see that 11𝑑 is equal to negative 121. And dividing both sides of this equation by 11 gives us 𝑑 is equal to negative 11. The common difference of the finite sequence is negative 11.

As we now know the first term and the common deference, all that is left to find is the last term of the finite sequence. In order to do this, we’ll use the fact that the twelfth-to-last term is negative 115. If we let π‘Ž sub 𝑙 be the last term, then the sequence can be written as shown. This means that π‘Ž sub 𝑙 minus one will be the second-to-last term or the penultimate term. π‘Ž sub 𝑙 minus two will be the third-to-last term. We can continue this pattern to show that π‘Ž sub 𝑙 minus 11 will be the twelfth-to-last term in the sequence.

Using the general formula once again, this is equal to π‘Ž sub one plus 𝑙 minus 11 minus one multiplied by 𝑑. Substituting in our values for π‘Ž sub 𝑙 minus 11, π‘Ž sub one, and 𝑑 gives us the equation negative 115 is equal to negative 82 plus 𝑙 minus 12 multiplied by negative 11. We can add 82 to both sides of this equation such that negative 33 is equal to 𝑙 minus 12 multiplied by negative 11. Dividing both sides of this equation by negative 11, we have three is equal to 𝑙 minus 12. We can then add 12 to both sides of this equation such that 𝑙 is equal to 15.

We can therefore conclude that the sequence has 15 terms. This fifteenth or last term, π‘Ž sub 15, is equal to negative 82 plus 15 minus one multiplied by negative 11. 15 minus one is 14. And multiplying this by negative 11 gives us negative 154. Negative 82 minus 154 is equal to negative 236. After clearing some space, we can summarize what we have found. Our sequence has a common difference of negative 11, a first term of negative 82, and a last term of negative 236. The finite arithmetic sequence π‘Ž sub 𝑛 contains the terms negative 82, negative 93, negative 104, and so on, with a last term of negative 236.

In our next example, we will solve the problem using simultaneous equations.

Find the arithmetic sequence in which the sum of the first and third terms equals negative 142 and the sum of its third and fourth terms equals negative 151.

In this question, we are given two pieces of information about our sequence. Firstly, the sum of the first and third terms equals negative 142. Secondly, the sum of the third and fourth terms equals negative 151. As we are dealing with an arithmetic sequence, we begin by recalling that the 𝑛th term, π‘Ž sub 𝑛, is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑, where π‘Ž sub one is the first term of the sequence and 𝑑 is the common difference.

In this question, we need to consider the first, third, and fourth terms of the sequence. The first term is equal to π‘Ž sub one. The third term is equal to π‘Ž sub one plus two 𝑑. And the fourth term is equal to π‘Ž sub one plus three 𝑑. This means that we have two equations. π‘Ž sub one plus π‘Ž sub one plus two 𝑑 is equal to negative 142. And π‘Ž sub one plus two 𝑑 plus π‘Ž sub one plus three 𝑑 is equal to negative 151.

The first equation simplifies to two π‘Ž sub one plus two 𝑑 is equal to negative 142. We will call this equation one. The second equation simplifies to two π‘Ž sub one plus five 𝑑 is equal to negative 151. We will label this equation two. We now have a pair of simultaneous equations that we can solve either by substitution or elimination. In this question, we will solve by elimination by subtracting equation one from equation two. Two π‘Ž sub one minus two π‘Ž sub one is zero, and five 𝑑 minus two 𝑑 is equal to three 𝑑.

On the right-hand side, subtracting negative 142 from negative 151 gives us negative nine. As three 𝑑 is equal to negative nine, we can divide both sides of our equation by three, giving us 𝑑 is equal to negative three. The common difference of the arithmetic sequence is negative three. We can now substitute this value of 𝑑 back in to equation one or equation two. Substituting 𝑑 equals negative three into equation one gives us two π‘Ž sub one plus two multiplied by negative three is equal to negative 142. This simplifies to two π‘Ž sub one minus six equals negative 142. We can then add six to both sides of this equation. Finally, dividing both sides by two gives us π‘Ž sub one is equal to negative 68.

The first term of our arithmetic sequence is negative 68, and the common difference is negative three. We can therefore conclude that our arithmetic sequence contains the values negative 68, negative 71, negative 74, and so on.

In our final question, we will find an arithmetic sequence given the sum and product of its terms.

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 multiplied by π‘Ž sub five is equal to 140.

In this question, we are told that the sum of the second and fourth terms of an arithmetic sequence equals negative 28 and the product of the third and fifth terms equals 140. We begin by recalling that the 𝑛th term of any arithmetic sequence, written π‘Ž sub 𝑛, is equal to π‘Ž sub one, the first term, plus 𝑛 minus one multiplied by 𝑑, the common difference. We can therefore set up two equations for the information given. Firstly, as the second term is π‘Ž sub one plus 𝑑 and the fourth term is π‘Ž sub one plus three 𝑑, we have the equation π‘Ž sub one plus 𝑑 plus π‘Ž sub one plus three 𝑑 is equal to negative 28. This simplifies to two π‘Ž sub one plus four 𝑑 is equal to negative 28.

At this stage, we can divide both sides of our equation by two. π‘Ž sub one plus two 𝑑 is therefore equal to negative 14. Subtracting two 𝑑 from both sides of this equation, we have π‘Ž sub one is equal to negative 14 minus two 𝑑. We will label this equation one. Next, we will consider the fact that π‘Ž sub three multiplied by π‘Ž sub five is 140. As the third term, π‘Ž sub three, is equal to π‘Ž sub one plus two 𝑑 and the fifth term, π‘Ž sub five, is equal to π‘Ž sub one plus four 𝑑, the product of these two expressions is equal to 140.

We could distribute the parentheses here. However, an alternative method would be to substitute the value of π‘Ž sub one into this equation. Replacing π‘Ž sub one with negative 14 minus two 𝑑, we have negative 14 minus two 𝑑 plus two 𝑑 multiplied by negative 14 minus two 𝑑 plus four 𝑑 is equal to 140. Negative two 𝑑 plus two 𝑑 is equal to zero. Negative two 𝑑 plus four 𝑑 is equal to two 𝑑. So we are left with negative 14 multiplied by negative 14 plus two 𝑑 is equal to 140.

We can divide both sides of this equation by negative 14 such that negative 14 plus two 𝑑 is equal to negative 10. Adding 14 to both sides, we see that two 𝑑 is equal to four. And finally, dividing both sides of this equation by two, we see that 𝑑 is equal to two. We can now substitute this value for the common difference into equation one to calculate the first term, π‘Ž sub one. π‘Ž sub one is equal to negative 14 minus two multiplied by two. This is equal to negative 18. Since the first term in our sequence is negative 18 and the common difference is two, the arithmetic sequence is negative 18, negative 16, negative 14, and so on.

We will now summarize the key points from this video. An arithmetic sequence is a sequence in which the difference between any two consecutive terms is constant. This is called the common deference. The 𝑛th term in an arithmetic sequence with first term π‘Ž sub one and common difference 𝑑 can be found using the formula π‘Ž sub 𝑛 is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑. We saw in this video that we can use this formula to construct an expression for any term in our arithmetic sequence in terms of the initial or first term and common difference. If we can then create two equations in these variables, we can solve them as simultaneous equations to calculate the values of π‘Ž sub one and 𝑑.

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