If `a_1,a_2,….a_n` are in H.P then `(a_1)/(a_2+,a_3,…,a_n),(a_2)/(a_1+a_3+….+a_n),…,(a_n)/(a_1+a_2+….+a_(n-1))`
are in

Options

A. A.P
B. G.P
C. H.P
D. none of these

Explanation

To determine if the sequence \(\left( \frac{a_1}{a_2 + a_3 + \ldots + a_n}, \frac{a_2}{a_1 + a_3 + \ldots + a_n}, \ldots, \frac{a_n}{a_1 + a_2 + \ldots + a_{n-1}} \right)\) is in Harmonic Progression (H.P), we start by recalling that if \(a_1, a_2, \ldots, a_n\) are in H.P, then their reciprocals \( \frac{1}{a_1}, \frac{1}{a_2}, \ldots, \frac{1}{a_n} \) are in Arithmetic Progression (A.P).

### Step-by-step Solution:

1. **Understanding H.P and A.P**:
Since \(a_1, a_2, \ldots, a_n\) are in H.P, we know that:
\[ \frac{1}{a_1}, \frac{1}{a_2}, \ldots, \frac{1}{a_n} \text{ are in A.P.} \]

2. **Expressing the Terms**:
The terms we need to analyze are:
\[ b_1 = \frac{a_1}{a_2 + a_3 + \ldots + a_n}, \quad b_2 = \frac{a_2}{a_1 + a_3 + \ldots + a_n}, \quad \ldots, \quad b_n = \frac{a_n}{a_1 + a_2 + \ldots + a_{n-1}}. \]

3. **Finding a Common Denominator**:
We can express the denominators in a more manageable form. Let:
\[ S = a_1 + a_2 + \ldots + a_n. \]
Then the terms can be rewritten as:
\[ b_1 = \frac{a_1}{S - a_1}, \quad b_2 = \frac{a_2}{S - a_2}, \quad \ldots, \quad b_n = \frac{a_n}{S - a_n}. \]

4. **Identifying the A.P**:
Since \( \frac{1}{a_1}, \frac{1}{a_2}, \ldots, \frac{1}{a_n} \) are in A.P, we can express the condition for A.P:
\[ 2 \cdot \frac{1}{a_k} = \frac{1}{a_{k-1}} + \frac{1}{a_{k+1}} \quad \text{for } k = 2, 3, \ldots, n-1. \]

5. **Reciprocal of the Terms**:
The terms \(b_k\) can be expressed in terms of their reciprocals:
\[ \frac{1}{b_k} = \frac{S - a_k}{a_k}. \]
Thus, we can see that if \(b_k\) are in H.P, then \(\frac{1}{b_k}\) must be in A.P.

6. **Conclusion**:
Since we have shown that the sequence \(\left( \frac{1}{b_1}, \frac{1}{b_2}, \ldots, \frac{1}{b_n} \right)\) is in A.P, it follows that the sequence \(b_1, b_2, \ldots, b_n\) is in H.P.

### Final Answer:
Thus, the sequence \(\left( \frac{a_1}{a_2 + a_3 + \ldots + a_n}, \frac{a_2}{a_1 + a_3 + \ldots + a_n}, \ldots, \frac{a_n}{a_1 + a_2 + \ldots + a_{n-1}} \right)\) is in **Harmonic Progression (H.P)**.