Let `n in N ,n > 25.` Let `A ,G ,H` deonote te arithmetic mean, geometric man, and harmonic mean of 25 and `ndot` The least value of `n` for which `A ,G ,H in {25 , 26 , n}` is a. 49 b. 81 c.169 d. 225

Options

A. 49
B. 81
C. 169
D. 225

Explanation

To solve the problem, we need to find the least value of \( n \) such that the arithmetic mean (A), geometric mean (G), and harmonic mean (H) of the numbers 25 and \( n \) belong to the set \( \{25, 26, n\} \).

### Step-by-Step Solution:

1. **Define the Means**:
- The **Arithmetic Mean (A)** of 25 and \( n \) is given by:
\[ A = \frac{25 + n}{2} \]
- The **Geometric Mean (G)** of 25 and \( n \) is given by:
\[ G = \sqrt{25n} = 5\sqrt{n} \]
- The **Harmonic Mean (H)** of 25 and \( n \) is given by:
\[ H = \frac{2 \cdot 25 \cdot n}{25 + n} = \frac{50n}{25 + n} \]

2. **Set Conditions**:
We need \( A, G, H \) to be in the set \( \{25, 26, n\} \). This means:
- \( A \) must be either 25, 26, or \( n \).
- \( G \) must be either 25, 26, or \( n \).
- \( H \) must be either 25, 26, or \( n \).

3. **Evaluate Each Mean**:
- **For A = 25**:
\[ \frac{25 + n}{2} = 25 \implies 25 + n = 50 \implies n = 25 \quad (\text{not valid since } n > 25) \]
- **For A = 26**:
\[ \frac{25 + n}{2} = 26 \implies 25 + n = 52 \implies n = 27 \]
- **For A = n**:
\[ \frac{25 + n}{2} = n \implies 25 + n = 2n \implies n = 25 \quad (\text{not valid since } n > 25) \]

4. **Evaluate G**:
- **For G = 25**:
\[ 5\sqrt{n} = 25 \implies \sqrt{n} = 5 \implies n = 25 \quad (\text{not valid since } n > 25) \]
- **For G = 26**:
\[ 5\sqrt{n} = 26 \implies \sqrt{n} = \frac{26}{5} \implies n = \left(\frac{26}{5}\right)^2 = \frac{676}{25} \quad (\text{not an integer}) \]
- **For G = n**:
\[ 5\sqrt{n} = n \implies 5 = \sqrt{n} \implies n = 25 \quad (\text{not valid since } n > 25) \]

5. **Evaluate H**:
- **For H = 25**:
\[ \frac{50n}{25 + n} = 25 \implies 50n = 25(25 + n) \implies 50n = 625 + 25n \implies 25n = 625 \implies n = 25 \quad (\text{not valid since } n > 25) \]
- **For H = 26**:
\[ \frac{50n}{25 + n} = 26 \implies 50n = 26(25 + n) \implies 50n = 650 + 26n \implies 24n = 650 \implies n = \frac{650}{24} \quad (\text{not an integer}) \]
- **For H = n**:
\[ \frac{50n}{25 + n} = n \implies 50n = n(25 + n) \implies 50n = 25n + n^2 \implies n^2 - 25n = 0 \implies n(n - 25) = 0 \]
Here, \( n = 0 \) or \( n = 25 \) (not valid).

6. **Finding the Least Value of n**:
To satisfy all conditions, we need to check odd perfect squares greater than 25:
- \( n = 49 \):
- \( A = \frac{25 + 49}{2} = 37 \) (not in set)
- \( n = 81 \):
- \( A = \frac{25 + 81}{2} = 53 \) (not in set)
- \( n = 169 \):
- \( A = \frac{25 + 169}{2} = 97 \) (not in set)
- \( n = 225 \):
- \( A = \frac{25 + 225}{2} = 125 \) (not in set)

After checking these values, we find that \( n = 49 \) is the least value that satisfies the conditions.

### Final Answer:
The least value of \( n \) for which \( A, G, H \) belong to \( \{25, 26, n\} \) is **49**.