What is the average of all multiples of 10 from 2 to 198?
100
Calculating the Average of Multiples of 10
The question asks us to find the average of all numbers that are multiples of 10 and fall within the range from 2 to 198.
Identifying the Multiples of 10
First, let's list the multiples of 10 that are greater than or equal to 2 and less than or equal to 198. A multiple of 10 must be divisible by 10. The multiples of 10 are 10, 20, 30, 40, and so on.
In the given range [2, 198], the multiples of 10 are:
- The smallest multiple of 10 greater than or equal to 2 is 10.
- The largest multiple of 10 less than or equal to 198 is 190.
So, the list of numbers is 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190.
Understanding Arithmetic Progression and Average
The list of multiples of 10 (10, 20, 30, ..., 190) forms an arithmetic progression (AP) because the difference between consecutive terms is constant (10). For any arithmetic progression, the average can be calculated easily if you know the first term and the last term.
The formula for the average of an arithmetic progression is:
\[ \text{Average} = \frac{\text{First Term} + \text{Last Term}}{2} \]
Calculating the Average of the Multiples
In our list of multiples of 10 from 2 to 198:
- The first term is 10.
- The last term is 190.
Now, we can use the average formula for an arithmetic progression:
\[ \text{Average} = \frac{10 + 190}{2} \]
\[ \text{Average} = \frac{200}{2} \]
\[ \text{Average} = 100 \]
Therefore, the average of all multiples of 10 from 2 to 198 is 100.
| Term | Value |
|---|---|
| First Multiple (\(\ge 2\)) | 10 |
| Last Multiple (\(\le 198\)) | 190 |
| Sum of First and Last | \(10 + 190 = 200\) |
| Average | \(200 / 2 = 100\) |
Revision Table: Average of Multiples
| Concept | Description |
|---|---|
| Multiples of 10 | Numbers divisible by 10 (e.g., 10, 20, 30...). |
| Range | The specified interval [2, 198]. |
| Arithmetic Progression (AP) | A sequence where the difference between consecutive terms is constant. Multiples of a number form an AP. |
| Average of AP | Can be found using \(\frac{\text{First Term} + \text{Last Term}}{2}\). |
Additional Information: Finding Number of Terms
While we didn't need the number of terms to find the average using the AP property, we can calculate it. The terms are 10, 20, ..., 190. This can be written as \(10 \times 1, 10 \times 2, \dots, 10 \times 19\). So, the terms are \(10n\), where \(n\) ranges from 1 to 19. The number of terms is 19.
Alternatively, using the AP formula for the n-th term \(l = a + (n-1)d\):
\[ 190 = 10 + (n-1)10 \]
\[ 180 = (n-1)10 \]
\[ 18 = n-1 \]
\[ n = 19 \]
There are 19 multiples of 10 in the range from 2 to 198.
The general formula for the average of any set of numbers is the sum of the numbers divided by the count of the numbers. For an AP, the sum is \(S_n = \frac{n}{2}(a + l)\). The average would then be \(\frac{S_n}{n} = \frac{\frac{n}{2}(a+l)}{n} = \frac{a+l}{2}\), which confirms the shortcut used.
Important Questions from Numerical Computation
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