I​f the 1st number in a series of consecutive odd numbers is 8 less than the last number in the series. How many numbers are there in the series? 

Let the series of consecutive odd numbers be \( x, x+2, x+4, \dots \). The first number is \( x \) and the last number is the \( n \)-th term, which is \( x + 2(n-1) \).

According to the problem, the first number is 8 less than the last number, so we have the equation:

\[
x + 8 = x + 2(n-1)
\]

Now, solve for \( n \):

\[
x + 8 = x + 2n - 2
\]

Subtract \( x \) from both sides:

\[
8 = 2n - 2
\]

Add 2 to both sides:

\[
10 = 2n
\]

Now divide by 2:

\[
n = 5
\]

Thus, there are 5 numbers in the series.

Would you like more details or have any questions? Here are 5 related questions to further explore the topic:

1. How would the series change if the difference were more than 8?
2. What happens if the series includes both even and odd numbers?
3. Can you find the sum of the numbers in this series?
4. What would happen if the series consisted of consecutive even numbers instead?
5. How would the solution change if the series started with a specific number?

**Tip**: When dealing with arithmetic sequences, always focus on the relationship between the first and last terms to set up equations easily.

I​f the 1st number in a series of consecutive odd numbers is 8 less than the last number in the series. How many numbers are there in the series?

Discover how to solve a problem involving consecutive odd numbers where the first number is 8 less than the last. This solution uses algebraic principles and arithmetic sequence formulas, perfect for Grades 6-8 students learning about sequences.

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