It `T_(k)` denotes the `k ^(th)` term of an H.P. from the bgegining and `(T_(2))/(T_(6)) =9,` then `(T_(10))/(T_(4))` equals :

Options

A. `17/5`
B. `5/17`
C. `7/19`
D. `19/7`

Answer

To solve the problem step by step, we need to use the properties of Harmonic Progression (H.P.) and the relationship between H.P. and Arithmetic Progression (A.P.).

### Step 1: Understanding the relationship between H.P. and A.P.
If \( T_k \) denotes the \( k^{th} \) term of an H.P., then the reciprocals of these terms, \( \frac{1}{T_k} \), will form an A.P.

Let the first term of this A.P. be \( a \) and the common difference be \( d \). Therefore, we can express the terms as:
- \( \frac{1}{T_1} = a \)
- \( \frac{1}{T_2} = a + d \)
- \( \frac{1}{T_3} = a + 2d \)
- \( \frac{1}{T_4} = a + 3d \)
- \( \frac{1}{T_5} = a + 4d \)
- \( \frac{1}{T_6} = a + 5d \)
- \( \frac{1}{T_{10}} = a + 9d \)

### Step 2: Using the given condition \( \frac{T_2}{T_6} = 9 \)
From the given condition, we can express it in terms of the A.P.:
\[ \frac{T_2}{T_6} = \frac{\frac{1}{\frac{1}{T_2}}}{\frac{1}{\frac{1}{T_6}}} = \frac{\frac{1}{a + d}}{\frac{1}{a + 5d}} = \frac{a + 5d}{a + d} \]
Setting this equal to 9, we have:
\[ \frac{a + 5d}{a + d} = 9 \]

### Step 3: Cross-multiplying to solve for \( a \) and \( d \)
Cross-multiplying gives:
\[ a + 5d = 9(a + d) \]
Expanding the right side:
\[ a + 5d = 9a + 9d \]
Rearranging the equation:
\[ 5d - 9d = 9a - a \]
This simplifies to:
\[ -4d = 8a \quad \Rightarrow \quad d = -2a \]

### Step 4: Finding \( \frac{T_{10}}{T_4} \)
Now we need to find \( \frac{T_{10}}{T_4} \):
\[ \frac{T_{10}}{T_4} = \frac{\frac{1}{\frac{1}{T_{10}}}}{\frac{1}{\frac{1}{T_4}}} = \frac{\frac{1}{a + 9d}}{\frac{1}{a + 3d}} = \frac{a + 3d}{a + 9d} \]

### Step 5: Substituting \( d = -2a \)
Substituting \( d \) into the expression:
\[ \frac{a + 3(-2a)}{a + 9(-2a)} = \frac{a - 6a}{a - 18a} = \frac{-5a}{-17a} = \frac{5}{17} \]

### Final Answer
Thus, we find:
\[ \frac{T_{10}}{T_4} = \frac{5}{17} \]