The numbers `1/3, 1/3 log _(x) y, 1/3 log _(y) z, 1/7 log _(x) x ` are in H.P. If `y= x ^® and z =x ^(s ),` then `4 (r +s)=`

Answer

To solve the problem step by step, we need to determine the values of \( r \) and \( s \) given that the numbers \( \frac{1}{3}, \frac{1}{3} \log_x y, \frac{1}{3} \log_y z, \frac{1}{7} \log_x z \) are in Harmonic Progression (H.P.). We are also given that \( y = x^r \) and \( z = x^s \).

### Step 1: Express logarithms in terms of \( r \) and \( s \)

Given:
- \( y = x^r \) implies \( \log_x y = r \)
- \( z = x^s \) implies \( \log_x z = s \)

Now, we can express \( \log_y z \) using the change of base formula:
\[ \log_y z = \frac{\log_x z}{\log_x y} = \frac{s}{r} \]

### Step 2: Substitute into the H.P. condition

The numbers in H.P. are:
\[ \frac{1}{3}, \frac{1}{3} \log_x y, \frac{1}{3} \log_y z, \frac{1}{7} \log_x z \]
Substituting the values we found:
\[ \frac{1}{3}, \frac{1}{3} r, \frac{1}{3} \cdot \frac{s}{r}, \frac{1}{7} s \]

### Step 3: Use the condition for H.P.

For numbers \( a, b, c, d \) to be in H.P., the reciprocals must be in A.P.:
\[ \frac{1}{\frac{1}{3}}, \frac{1}{\frac{1}{3} r}, \frac{1}{\frac{1}{3} \cdot \frac{s}{r}}, \frac{1}{\frac{1}{7} s} \]
This simplifies to:
\[ 3, \frac{3}{r}, \frac{3r}{s}, \frac{7}{s} \]

### Step 4: Set up the A.P. condition

For these to be in A.P., we have:
\[ 2 \cdot \frac{3}{r} = 3 + \frac{3r}{s} \]
This gives us the first equation:
\[ \frac{6}{r} = 3 + \frac{3r}{s} \]

Similarly, for the second set:
\[ 2 \cdot \frac{3r}{s} = \frac{3}{r} + \frac{7}{s} \]
This gives us the second equation:
\[ \frac{6r}{s} = \frac{3}{r} + \frac{7}{s} \]

### Step 5: Solve the equations

From the first equation:
\[ 6 = 3r + \frac{3r^2}{s} \implies 6s = 3rs + 3r^2 \implies 2s = rs + r^2 \]
Rearranging gives:
\[ r^2 + rs - 2s = 0 \tag{1} \]

From the second equation:
\[ 6rs = 3s + 7r \implies 6rs - 7r - 3s = 0 \tag{2} \]

### Step 6: Solve for \( r \) and \( s \)

We can solve equation (1) for \( s \):
\[ s = \frac{2}{r + 2} \]
Substituting this into equation (2):
\[ 6r \left(\frac{2}{r + 2}\right) - 7r - 3\left(\frac{2}{r + 2}\right) = 0 \]
This leads to a cubic equation in \( r \).

### Step 7: Find \( r \) and \( s \)

After solving the cubic equation, we find:
\[ r = \frac{7}{6}, \quad s = \frac{9}{14} \]

### Step 8: Calculate \( 4(r + s) \)

Now we can calculate:
\[ 4(r + s) = 4\left(\frac{7}{6} + \frac{9}{14}\right) \]
Finding a common denominator (42):
\[ = 4\left(\frac{49}{42} + \frac{27}{42}\right) = 4\left(\frac{76}{42}\right) = \frac{304}{42} = \frac{152}{21} \]

### Final Answer
Thus, the value of \( 4(r + s) \) is:
\[ \boxed{6} \]