To solve the problem, we need to find the values of \( k \) in the series \( 21, 22, 23, \ldots, k-1, k \) such that both the arithmetic mean (AM) and geometric mean (GM) of the first and last terms exist in the series, and \( k \) is a three-digit number.
### Step-by-Step Solution:
1. **Identify the Series**:
The series is given as \( 21, 22, 23, \ldots, k-1, k \).
2. **Calculate the Arithmetic Mean (AM)**:
The arithmetic mean of the first term (21) and the last term (k) is given by:
\[
AM = \frac{21 + k}{2}
\]
For the AM to be an integer, \( 21 + k \) must be even. Since 21 is odd, \( k \) must also be odd (because odd + odd = even).
3. **Calculate the Geometric Mean (GM)**:
The geometric mean of the first term (21) and the last term (k) is given by:
\[
GM = \sqrt{21 \cdot k}
\]
For the GM to be an integer, \( 21 \cdot k \) must be a perfect square.
4. **Express \( k \)**:
Since \( k \) must be odd, we can express \( k \) as:
\[
k = 21n^2
\]
where \( n \) is an integer, such that \( 21n^2 \) is a three-digit number.
5. **Determine the Range for \( n \)**:
We need \( k \) to be a three-digit number:
\[
100 \leq 21n^2 < 1000
\]
Dividing through by 21 gives:
\[
\frac{100}{21} \leq n^2 < \frac{1000}{21}
\]
This simplifies to:
\[
4.76 \leq n^2 < 47.62
\]
Taking square roots:
\[
2.18 \leq n < 6.9
\]
Therefore, \( n \) can take integer values \( 3, 4, 5, 6 \).
6. **Calculate Possible Values of \( k \)**:
- For \( n = 3 \):
\[
k = 21 \cdot 3^2 = 21 \cdot 9 = 189
\]
- For \( n = 4 \):
\[
k = 21 \cdot 4^2 = 21 \cdot 16 = 336
\]
- For \( n = 5 \):
\[
k = 21 \cdot 5^2 = 21 \cdot 25 = 525
\]
- For \( n = 6 \):
\[
k = 21 \cdot 6^2 = 21 \cdot 36 = 756
\]
7. **Count the Valid Values of \( k \)**:
The valid values of \( k \) that are three-digit odd numbers are:
- \( 189 \)
- \( 525 \)
- \( 756 \)
Thus, there are **three valid values of \( k \)**.
### Conclusion:
The values of \( k \) that satisfy the conditions are \( 189, 525, \) and \( 756 \). Therefore, \( k \) can attain **3 values**.
