For the series `21,22,23 ,…. K - 1 , k ` the A.M and G.M of the first and last numbers exist in the given series . If 'k' is a three digit number , then 'k' can attian ______ values .

Answer

To solve the problem, we need to find the values of \( k \) in the series \( 21, 22, 23, \ldots, k-1, k \) such that the arithmetic mean (A.M) and geometric mean (G.M) of the first and last terms exist in the series. We also know that \( k \) is a three-digit number.

### Step-by-Step Solution:

1. **Identify the Series**:
The series is \( 21, 22, 23, \ldots, k-1, k \). This is an arithmetic progression (AP) with a common difference of 1.

2. **Calculate the Arithmetic Mean (A.M)**:
The A.M of the first term (21) and the last term (k) is given by:
\[ \text{A.M} = \frac{21 + k}{2} \]
For A.M to be an integer, \( 21 + k \) must be even. Since 21 is odd, \( k \) must also be odd for their sum to be even.

3. **Calculate the Geometric Mean (G.M)**:
The G.M of the first term (21) and the last term (k) is given by:
\[ \text{G.M} = \sqrt{21 \cdot k} \]
For G.M to be an integer, \( 21 \cdot k \) must be a perfect square. Since 21 can be factored as \( 3 \cdot 7 \), \( k \) must include the factors such that \( 21 \cdot k \) becomes a perfect square.

4. **Express \( k \)**:
Let \( k = 21n^2 \) where \( n \) is a positive integer. This ensures that \( 21 \cdot k \) is a perfect square because:
\[ 21 \cdot k = 21 \cdot (21n^2) = 21^2 \cdot n^2 \]
which is a perfect square.

5. **Determine the Range for \( k \)**:
Since \( k \) is a three-digit number:
\[ 100 \leq 21n^2 < 1000 \]
Dividing the entire inequality by 21 gives:
\[ \frac{100}{21} \leq n^2 < \frac{1000}{21} \]
This simplifies to:
\[ 4.76 \leq n^2 < 47.62 \]
Therefore, \( n^2 \) can take integer values from 5 to 47.

6. **Find Possible Values of \( n \)**:
The integer values of \( n^2 \) are 5, 6, 7, ..., 47. The corresponding integer values of \( n \) can be calculated:
- \( n = 3 \) gives \( n^2 = 9 \)
- \( n = 4 \) gives \( n^2 = 16 \)
- \( n = 5 \) gives \( n^2 = 25 \)
- \( n = 6 \) gives \( n^2 = 36 \)
- \( n = 7 \) gives \( n^2 = 49 \)

7. **Calculate Values of \( k \)**:
- For \( n = 3 \): \( k = 21 \cdot 9 = 189 \)
- For \( n = 4 \): \( k = 21 \cdot 16 = 336 \)
- For \( n = 5 \): \( k = 21 \cdot 25 = 525 \)
- For \( n = 6 \): \( k = 21 \cdot 36 = 756 \)
- For \( n = 7 \): \( k = 21 \cdot 49 = 1029 \) (not a three-digit number)

Thus, the valid values of \( k \) are \( 189, 336, 525, 756 \).

### Final Answer:
The number of possible values of \( k \) is **4**.