Lesson Explainer: Arithmetic Mean Mathematics

In this explainer, we will learn how to find the arithmetic means for any two nonconsecutive terms in an arithmetic sequence.

The concept of finding the arithmetic mean of two or more numbers should be familiar; we use this measurement to find the average of a set of data.

Definition: Arithmetic Mean of Two Numbers

The arithmetic mean of a pair of numbers, 𝑥 and 𝑥, is given by 𝑥+𝑥2.

Let us begin by demonstrating an application of this formula.

Example 1: Finding Two Unknowns Using Information about Their Arithmetic Mean

If the arithmetic mean between 𝑎 and 𝑏 is 9 and the arithmetic mean between 7𝑎 and 5𝑏 is 15, then 𝑏𝑎=.

Answer

Recall that the arithmetic mean of a pair of numbers, 𝑥 and 𝑥, is given by 𝑥+𝑥2.

We are given that the arithmetic mean of 𝑎 and 𝑏 is 9, so 𝑎+𝑏2=9𝑎+𝑏=18.

Similarly, the arithmetic mean of 7𝑎 and 5𝑏 is 15, so 7𝑎+5𝑏2=157𝑎+5𝑏=30.

To solve these simultaneous equations by elimination, we multiply the first equation by 5: 5𝑎5𝑏=90.

Adding this equation to the second equation will eliminate the variable 𝑏: 5𝑎5𝑏=907𝑎+5𝑏=302𝑎=60,𝑎=30.+

Since 𝑎+𝑏=18, 30+𝑏=18𝑏=48.

Remember, our goal is to calculate the value of 𝑏𝑎. Substituting 𝑎=30 and 𝑏=48 into this expression gives 48(30)=78.

Therefore, 𝑏𝑎=78.

We can generalize the idea of an arithmetic mean to the terms in an arithmetic sequence.

Definition: 𝑛 Arithmetic Means

Given a pair of numbers 𝑎 and 𝑏, the 𝑛 arithmetic means between 𝑎 and 𝑏 are the values in an arithmetic sequence from 𝑎 to 𝑏 with exactly 𝑛 terms in between.

Consider the simple arithmetic sequence (4,8,12,16,20,24). In this case, the terms (8,12,16,20) are the 4 arithmetic means between 4 and 24.

In particular, the arithmetic mean of the 2nd and 4th terms is given by 𝑇+𝑇2=8+162=12=𝑇.

The third term is the arithmetic mean of the 2nd and 4th terms in an arithmetic sequence. In fact, if we take any three consecutive terms of an arithmetic sequence, the middle term is the arithmetic mean between the other two terms. We generalize this concept when we define the terms between any two terms of an arithmetic sequence to be arithmetic means.

We notice the relationship that exists between the term number and the order of the arithmetic means: The 1st mean is the 2nd term in an arithmetic sequence, the 4th arithmetic mean is the 5th term in the sequence, and so on. In general, the 𝑛th arithmetic mean is the (𝑛+1)th term in the arithmetic sequence.

We will now demonstrate how to use this definition to find arithmetic means between a pair of numbers.

Example 2: Finding a Certain Number of Arithmetic Means between Two Numbers

Find 5 arithmetic means between 7 and 19.

Answer

To find the 5 arithmetic means between 7 and 19, we need to identify the arithmetic sequence from 7 to 19 with exactly 5 terms in between.

We recall that the general term of an arithmetic sequence with first term 𝑇 and common difference 𝑑 is 𝑇=𝑇+𝑑(𝑛1).

In order to find the terms between 7 and 19, the first step is to find the common difference, 𝑑. Since there are 5 terms in between 7 and 19, we know that 7 is the first term and 19 is the 7th term of the arithmetic sequence. Substituting 𝑇=7, 𝑇=19, and 𝑛=7 into the formula for the 𝑛th term, 19=7+𝑑(71)19=7+6𝑑12=6𝑑𝑑=2.

Since the first term is 7 and the common difference is 2, the arithmetic sequence is 7, 9, 11, 13, 15, 17, 19. We note that there are exactly 5 terms between 7 and 19, as expected.

Hence, the 5 arithmetic means between 7 and 19 are (9,11,13,15,17).

In the next example, we will demonstrate how to use information about arithmetic means in sequences to set up simultaneous equations. The solutions to these simultaneous equations will then allow us to define an arithmetic sequence.

Example 3: Finding an Arithmetic Sequence Using Information about the Arithmetic Means

If the sum of the second mean and the fourth mean from an arithmetic sequence equals 16 and the seventh mean is more than the third mean by 8, then the sequence is .

Answer

The general term of an arithmetic sequence with first term 𝑇 and common difference 𝑑 is given by 𝑇=𝑇+𝑑(𝑛1). Since an arithmetic mean does not include the first term of the sequence, the second mean actually corresponds to the third term of the sequence, 𝑇: 𝑇=𝑇+2𝑑.

Similarly, we can create expressions in terms of 𝑇 and 𝑑 for the third mean (𝑇), fourth mean (𝑇), and seventh mean (𝑇), giving 𝑇=𝑇+3𝑑,𝑇=𝑇+4𝑑,𝑇=𝑇+7𝑑.

We now create a pair of simultaneous equations using the given information. Since the sum of the second and fourth mean is 16, this gives us the equation 𝑇+𝑇=16, which in turn gives (𝑇+2𝑑)+(𝑇+4𝑑)=162𝑇+6𝑑=16.

Since the seventh mean is 8 more than the third mean, we form a second equation using 𝑇+8=𝑇: (𝑇+3𝑑)+8=(𝑇+7𝑑)3𝑑+8=7𝑑8=4𝑑2=𝑑.

Substituting 𝑑=2 into the first equation, 2𝑇+6(2)=162𝑇+12=162𝑇=4𝑇=2.

Hence, the first term of this sequence is 2. The arithmetic sequence with a first term of 2 and a common difference of 2 is 2,4,6.

In our next example, we will use given information about arithmetic means to identify the number of arithmetic means between two given values.

Example 4: Finding the Number of Arithmetic Means Inserted between Two Numbers given the Sum of Two Means

Find the number of arithmetic means inserted between 8 and 238 given the sum of the second and the sixth means is 96.

Answer

We recall that the number of arithmetic means between two values is the number of terms between the given terms in an arithmetic sequence. Recall that the general form of the arithmetic sequence is 𝑇=𝑇+𝑑(𝑛1), where 𝑇 is the first term and 𝑑 is the common difference. We also know that the first term of the sequence is not an arithmetic mean, so the 𝑛th mean is the (𝑛+1)th term, 𝑇. Hence, the second mean is the third term, 𝑇, and the sixth mean is the seventh term, 𝑇. The second and sixth means can therefore be written in terms of the first term, 8, and the common difference, 𝑑, 𝑇=8+2𝑑,𝑇=8+6𝑑.

Since the sum of the second and sixth mean is 96, (8+2𝑑)+(8+6𝑑)=9616+8𝑑=968𝑑=80𝑑=10.

To find the number of terms between 8 and 238, we can substitute 𝑇=8 and 𝑑=10 into the formula for the 𝑛th term of an arithmetic sequence: 𝑇=𝑇+𝑑(𝑛1).

Since the last term of the sequence is 238, we can substitute 𝑇=238: 238=8+10(𝑛1)238=8+10𝑛10240=10𝑛𝑛=24.

There are 24 terms in the sequence, which means there are 242=22 terms between 8 and 238.

Under these conditions, the number of arithmetic means between 8 and 238 is 22.

In our final example, we will again need to calculate the number of arithmetic means between two given values.

Example 5: Finding the Number of Arithmetic Means Inserted between Two Numbers Using Information about the Sum of Arithmetic Means

Find the number of arithmetic means inserted between 2 and 254 given the ratio between the sum of the first two means and the sum of the last two means is 11245.

Answer

The number of arithmetic means between two values is the number of terms between the first and final terms in an arithmetic sequence, where the general form of an arithmetic sequence is 𝑇=𝑇+𝑑(𝑛1). If we let 𝑇=2, we can write the first two means in terms of the first term, 𝑇, and the common difference, 𝑑. The first mean is the second term in the sequence. Therefore, 𝑇=2+𝑑.

The second mean is the third term in sequence. Therefore, 𝑇=2+2𝑑.

We know the ratio of the sum of the first two means and the sum of the last two means. We can combine the two expressions for the first and second mean to give the following expression for the sum of the first two means: 𝑇+𝑇=(2+𝑑)+(2+2𝑑)=4+3𝑑.

We follow a similar process for the final two means. However, we do not know the term number of the final two means, so we write these in terms of the last term and the common difference. The second-to-last term will be the last arithmetic mean. It can also be written in terms of its relationship to the final term as 254𝑑.

The second-to-last mean is the term before the final mean. We can write that expression as (254𝑑)𝑑=2542𝑑.

The expression for the sum of the final two means is (254𝑑)+(2542𝑑)=5083𝑑.

We use these expressions to set up the equivalent ratio: 4+3𝑑5083𝑑=11245.

To solve for 𝑑, we cross multiply and then distribute the parentheses which gives 254(4+3𝑑)=11(5083𝑑)980+735𝑑=558833768𝑑=4608𝑑=6.

To find the number of terms between 2 and 254, we can substitute 𝑇=2 and 𝑑=6 into the formula for the 𝑛th term of an arithmetic sequence: 𝑇=𝑇+𝑑(𝑛1).

Since the last term of the sequence is 254, we substitute 𝑇=254: 254=2+6(𝑛1)254=2+6𝑛6258=6𝑛𝑛=43.

Since there are 43 terms in the sequence, the number of terms between 2 and 254 is 432=41.

Hence, the number of arithmetic means between 2 and 254 is 41.

Let us finish by recapping a few important concepts from the explainer.

Key Points

  • Given a pair of numbers 𝑎 and 𝑏, the 𝑛 arithmetic means between 𝑎 and 𝑏 are the values in an arithmetic sequence from 𝑎 to 𝑏 with exactly 𝑛 terms in between.
  • In an arithmetic sequence of the general form 𝑇=𝑇+𝑑(𝑛1), the values from 𝑇 to 𝑇 are called arithmetic means. The 𝑘th arithmetic mean is the (𝑘+1)th term.
  • Since the arithmetic means are all of the terms in an arithmetic sequence except the first and final term, for a sequence with 𝑛 number of terms, there are 𝑛2 arithmetic means between 𝑇 and 𝑇.

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