What is the 10th number in the sequence of 3, 5/2, 2, 3/2, 1...

The given sequence is \(3, \frac{5}{2}, 2, \frac{3}{2}, 1, \ldots\). To find the 10th number in the sequence, we first need to determine the pattern governing the sequence.

Looking closely, we observe that each term decreases by \(\frac{1}{2}\) from the previous term:

\[
\begin{aligned}
&3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2},\\
&\frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2,\\
&2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2},\\
&\frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1.
\end{aligned}
\]

From this pattern, the \(n\)-th term of the sequence can be expressed as:
\[
a_n = 3 - \left( n-1 \right) \cdot \frac{1}{2}.
\]

To find the 10th term (\(a_{10}\)):
\[
\begin{aligned}
a_{10} &= 3 - (10-1) \cdot \frac{1}{2} \\
       &= 3 - 9 \cdot \frac{1}{2} \\
       &= 3 - \frac{9}{2} \\
       &= \frac{6}{2} - \frac{9}{2} \\
       &= \frac{6 - 9}{2} \\
       &= \frac{-3}{2} \\
       &= -\frac{3}{2}.
\end{aligned}
\]

Therefore, the 10th number in the sequence is \(-\frac{3}{2}\).

Would you like more details or have any questions?

Here are 8 related questions you might find interesting:

1. What is the 15th number in the sequence?
2. How can you generalize this sequence to an \(n\)-th term formula?
3. What is the sum of the first 10 terms of this sequence?
4. How does the sequence behave as \(n\) approaches infinity?
5. Can this sequence be considered an arithmetic progression?
6. What is the common difference of the sequence?
7. How would the sequence change if the first term was 4 instead of 3?
8. What is the 50th term of this sequence?

**Tip:** When identifying patterns in sequences, always check the difference between consecutive terms to see if there is a constant difference, which indicates an arithmetic sequence.

Learn how to find the 10th number in the sequence 3, 5/2, 2, 3/2, 1... by identifying the pattern and applying the formula for the n-th term of an arithmetic sequence. This problem involves sequences and arithmetic progressions, suitable for students in Grades 9-12.

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