Lesson Video: Using Arithmetic Sequence Formulas Mathematics

In this video, we will learn how to write explicit and recursive formulas for arithmetic sequences to find the value of the 𝑛th term in an arithmetic sequence and how to find a term’s order given its value.

14:50

Video Transcript

In this video, we will learn how to write explicit and recursive formulas for arithmetic sequences to find the value of the 𝑛th term in an arithmetic sequence and how to find a term’s order given its value. We will begin by recalling some key definitions of arithmetic sequences.

An arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant, for example, the sequence seven, 11, 15, 19, and so on. The difference between two consecutive terms in this sequence is four. This value is known as the common difference. The first term of an arithmetic sequence is usually denoted π‘Ž sub one, with the second term π‘Ž sub two, and so on. However, it is worth noting that occasionally the first term is denoted π‘Ž sub zero, the second term π‘Ž sub one, and so on.

The 𝑛th term, denoted π‘Ž sub 𝑛, of an arithmetic sequence with common difference 𝑑 and first term π‘Ž sub one is given by π‘Ž sub 𝑛 is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑. In our example, π‘Ž sub one is equal to seven, and the common difference 𝑑 is equal to four. This means that π‘Ž sub 𝑛 is equal to seven plus 𝑛 minus one multiplied by four. Distributing our parentheses, this becomes seven plus four 𝑛 minus four, which is equal to four 𝑛 plus three. We can then use this formula to calculate the value of any term in the sequence. For example, to calculate the tenth term, we would substitute 𝑛 is equal to 10. The term with order π‘Ž sub 10 is equal to four multiplied by 10 plus three, which is equal to 43.

We will now look at a question where we need to calculate a specific term of an arithmetic sequence.

Find π‘Ž sub 45, given the arithmetic sequence 18, 26, 34, and so on, up to 698, where 𝑛 is greater than or equal to one.

In this question, we’re given an arithmetic sequence that begins 18, 26, 34. As we are told that 𝑛 is greater than or equal to one, we will denote these first three terms as π‘Ž sub one, π‘Ž sub two, and π‘Ž sub three. We know that any sequence is arithmetic if there’s a common difference between consecutive terms. In this question, the common difference is eight, as 18 plus eight is 26, and 26 plus eight is 34.

We are asked to calculate the 45th term in the sequence. We know that the 𝑛th term of any arithmetic sequence with common difference 𝑑 and first term π‘Ž sub one is given by π‘Ž sub 𝑛 is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑. This means that π‘Ž sub 45 is equal to 18 plus 45 minus one multiplied by eight. This simplifies to 18 plus 44 multiplied by eight. Since 40 multiplied by eight is 320 and four multiplied by eight is 32, then 44 multiplied by eight is equal to 352. Adding 18 to this gives us a final answer of 370. The 45th term in the arithmetic sequence 18, 26, 34, and so on up to 698 is 370.

In our next question, we need to find the number of terms in a given arithmetic sequence.

Find the number of terms in the arithmetic sequence negative four, two, eight, and so on, up to 392.

We know that in order for a sequence to be arithmetic, it must have a common difference between consecutive terms. This is true in this question, as the difference between the first and second terms is the same as the difference between the second and third terms. We have a common difference equal to six.

We know that the 𝑛th term of any arithmetic sequence with common difference 𝑑 and first term π‘Ž sub one satisfies the equation π‘Ž sub 𝑛 is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑. We know that in this sequence, the last term is equal to 392. Substituting the values of π‘Ž sub one and 𝑑 into the right-hand side of our equation, we have negative four plus 𝑛 minus one multiplied by six. Distributing the parentheses, we have six 𝑛 minus six. And since the last term is equal to 392, we have negative four plus six 𝑛 minus six is equal to 392.

The left-hand side simplifies to six 𝑛 minus 10. And then we can add 10 to both sides of the equation such that six 𝑛 is equal to 402. Finally, dividing both sides of this equation by six, we have 𝑛 is equal to 402 over six, which is equal to 67. We can therefore conclude that there are 67 terms in the arithmetic sequence negative four, two, eight, and so on, up to 392, where the 67th term is 392.

In our next example, we will calculate the formula for the 𝑛th term of an arithmetic sequence.

Find, in terms of 𝑛, the general term of an arithmetic sequence whose sixth term is 46 and the sum of the third and tenth terms is 102.

We recall that a sequence is arithmetic if there is a common difference between any two consecutive terms. And the 𝑛th term, written π‘Ž sub 𝑛, of any arithmetic sequence is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑, where π‘Ž sub one is the first term and 𝑑 is the common difference.

In this question, we are told that the sixth term is 46. Therefore, π‘Ž sub six is equal to 46. We are also told that the sum of the third and tenth terms is 102. π‘Ž sub three plus π‘Ž sub 10 is equal to 102. Using the general equation, we have π‘Ž sub one plus six minus one multiplied by 𝑑 is equal to 46. This simplifies to π‘Ž sub one plus five 𝑑 equals 46. And as there are two unknowns, we will call this equation one.

The third term, π‘Ž sub three, is equal to π‘Ž sub one plus two 𝑑. And the tenth term, π‘Ž sub 10, is equal to π‘Ž sub one plus nine 𝑑. As these sum to give us 102, we have π‘Ž sub one plus two 𝑑 plus π‘Ž sub one plus nine 𝑑 is equal to 102. The left-hand side simplifies to two π‘Ž sub one plus 11𝑑. If we call this equation two, we now have a pair of simultaneous equations in the two unknowns π‘Ž sub one and 𝑑.

One way to solve these would be by elimination. Multiplying equation one by two gives us two π‘Ž sub one plus 10𝑑 is equal to 92. If we call this equation three, we can then subtract this from equation two. This leaves us with 𝑑 is equal to 10. Substituting this back into equation one gives us π‘Ž sub one plus five multiplied by 10 is equal to 46. This means that π‘Ž sub one plus 50 is equal to 46. Subtracting 50 from both sides of the equation gives us π‘Ž sub one is equal to negative four.

We now have values for the first term and common difference of our arithmetic sequence. And substituting them back into the general equation gives us π‘Ž sub 𝑛 is equal to negative four plus 𝑛 minus one multiplied by 10. By distributing the parentheses, this simplifies to negative four plus 10𝑛 minus 10, which is equal to 10𝑛 minus 14. The general term of an arithmetic sequence whose sixth term is 46 and sum of the third and tenth terms is 102 is 10𝑛 minus 14.

In our final example, we will find the order of a specific term in an arithmetic sequence.

Find the order of the term whose value is 112 in the sequence 17, 22, 27, 32, and so on.

We begin by recalling that the order of a term in a sequence is its position. In this question, we are given the sequence 17, 22, 27, 32, and so on. This means that the first term, π‘Ž sub one, is equal to 17. The second term, π‘Ž sub two, is 22. The third term, π‘Ž sub three, is 27, and so on. We need to find the order or position of the term whose value is 112.

We know that our sequence is arithmetic, as there is a common difference between consecutive terms. In any arithmetic sequence, the general 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 and 𝑑 is the common difference. We can see from our sequence that π‘Ž sub one is equal to 17 and the common difference 𝑑 is five. Substituting these values into the general equation gives us 17 plus 𝑛 minus one multiplied by five.

We want to find the value of 𝑛 such that this expression is equal to 112. Distributing the parentheses, we have 17 plus five 𝑛 minus five is equal to 112. The left-hand side simplifies to five 𝑛 plus 12. We can then subtract 12 from both sides such that five 𝑛 is equal to 100. Finally, dividing through by five gives us 𝑛 is equal to 20. And we can therefore conclude that the order of the term whose value is 112 is π‘Ž sub 20. It is the 20th term in the sequence 17, 22, 27, 32, and so on.

It is worth noting that all the even-number terms in this sequence end in two. This suggests that our answer of π‘Ž sub 20 is correct.

We will now summarize the key points from this video. We saw in this video that a sequence is a arithmetic if there is a common difference between any two consecutive terms. The 𝑛th term of an arithmetic sequence with common difference 𝑑 and first term π‘Ž sub one is given by π‘Ž sub 𝑛 is equal to π‘Ž sub one plus 𝑛 minus one multiplied by 𝑑. In this video, we used this equation to help calculate a specific term of an arithmetic sequence, in order to find the number of terms in a given sequence, or to calculate the formula for a specific arithmetic sequence.

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