The value of x + y + z is 15 if a, x, y, z, b are in A.P.
while the value of ` (1)/(x) + (1)/(y) + (1)/(z) + (1)/(z) is (5)/(3)` if a, x, y, z, b are
in H.P. the value of a and b are

Options

A. 9,1
B. 7,4
C. 8,2
D. `-1,3`

Answer

To solve the problem step by step, we need to find the values of \( a \) and \( b \) given the conditions about the arithmetic progression (A.P.) and harmonic progression (H.P.) of the numbers \( a, x, y, z, b \).

### Step 1: Set up the equations from the A.P. condition
Given that \( a, x, y, z, b \) are in A.P., we know that the average of the first and last terms equals the average of the middle terms. The sum of \( x + y + z = 15 \).

Using the formula for the sum of an A.P.:
\[ x + y + z = \frac{n}{2} \times (a + b) \]
where \( n = 5 \) (the number of terms). Thus, we have:
\[ 15 = \frac{5}{2} \times (a + b) \]
Multiplying both sides by 2:
\[ 30 = 5(a + b) \]
Dividing by 5:
\[ a + b = 6 \quad \text{(Equation 1)} \]

### Step 2: Set up the equations from the H.P. condition
Next, we consider the condition that \( a, x, y, z, b \) are in H.P. This means that \( \frac{1}{a}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{b} \) are in A.P.

Using the formula for the sum of an A.P. again, we have:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{2} \left( \frac{1}{a} + \frac{1}{b} \right) \]
We know from the problem statement that:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3} \]
So we equate:
\[ \frac{5}{3} = \frac{5}{2} \left( \frac{1}{a} + \frac{1}{b} \right) \]
Dividing both sides by \( \frac{5}{2} \):
\[ \frac{2}{3} = \frac{1}{a} + \frac{1}{b} \]
This can be rewritten as:
\[ \frac{1}{a} + \frac{1}{b} = \frac{2}{3} \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
From Equation 1, we have:
\[ b = 6 - a \]
Substituting \( b \) into Equation 2:
\[ \frac{1}{a} + \frac{1}{6 - a} = \frac{2}{3} \]
Finding a common denominator:
\[ \frac{(6 - a) + a}{a(6 - a)} = \frac{2}{3} \]
This simplifies to:
\[ \frac{6}{a(6 - a)} = \frac{2}{3} \]
Cross-multiplying gives:
\[ 18 = 2a(6 - a) \]
Expanding:
\[ 18 = 12a - 2a^2 \]
Rearranging into standard quadratic form:
\[ 2a^2 - 12a + 18 = 0 \]
Dividing through by 2:
\[ a^2 - 6a + 9 = 0 \]
Factoring:
\[ (a - 3)^2 = 0 \]
Thus, we find:
\[ a = 3 \]
Substituting back into Equation 1:
\[ b = 6 - 3 = 3 \]

### Final Answer
The values of \( a \) and \( b \) are:
\[ \boxed{(3, 3)} \]