If `a_(a), a _(2), a _(3),…., a_(n)` are in H.P. and `f (k)=sum _(r =1) ^(n) a_(r)-a_(k) ` then `(a_(1))/(f(1)), (a_(2))/(f (2)), (a_(3))/(f (n))` are in :

Options

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

Answer

To solve the problem, we need to analyze the given information and derive the required relationships step by step.

### Step 1: Understanding Harmonic Progression (H.P.)

Given that \( a_1, a_2, a_3, \ldots, a_n \) are in Harmonic Progression (H.P.), we know that their reciprocals \( \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots, \frac{1}{a_n} \) are in Arithmetic Progression (A.P.).

### Step 2: Define the Function \( f(k) \)

The function \( f(k) \) is defined as:
\[ f(k) = \sum_{r=1}^{n} a_r - a_k \]
This means:
- \( f(1) = a_2 + a_3 + \ldots + a_n \)
- \( f(2) = a_1 + a_3 + \ldots + a_n \)
- \( f(3) = a_1 + a_2 + a_4 + \ldots + a_n \)
- ...
- \( f(n) = a_1 + a_2 + \ldots + a_{n-1} \)

### Step 3: Express \( \frac{a_k}{f(k)} \)

We need to analyze the ratios:
\[ \frac{a_1}{f(1)}, \frac{a_2}{f(2)}, \frac{a_3}{f(3)}, \ldots, \frac{a_n}{f(n)} \]

### Step 4: Calculate Each \( f(k) \)

1. For \( f(1) \):
\[ f(1) = a_2 + a_3 + \ldots + a_n \]
Thus,
\[ \frac{a_1}{f(1)} = \frac{a_1}{a_2 + a_3 + \ldots + a_n} \]

2. For \( f(2) \):
\[ f(2) = a_1 + a_3 + \ldots + a_n \]
Thus,
\[ \frac{a_2}{f(2)} = \frac{a_2}{a_1 + a_3 + \ldots + a_n} \]

3. For \( f(3) \):
\[ f(3) = a_1 + a_2 + a_4 + \ldots + a_n \]
Thus,
\[ \frac{a_3}{f(3)} = \frac{a_3}{a_1 + a_2 + a_4 + \ldots + a_n} \]

### Step 5: Analyze the Ratios

Since \( a_1, a_2, \ldots, a_n \) are in H.P., their reciprocals are in A.P. This implies that the ratios \( \frac{a_k}{f(k)} \) are also structured in a way that they maintain a consistent relationship.

### Step 6: Conclusion

Since we have established that the ratios \( \frac{a_1}{f(1)}, \frac{a_2}{f(2)}, \frac{a_3}{f(3)}, \ldots, \frac{a_n}{f(n)} \) maintain the properties of A.P., we conclude that:
\[ \frac{a_1}{f(1)}, \frac{a_2}{f(2)}, \frac{a_3}{f(3)}, \ldots, \frac{a_n}{f(n)} \text{ are in Harmonic Progression (H.P.)} \]

Thus, the correct answer is that these terms are in H.P.