Suppose a, x, y, z and b are in A.P. where x + y + z = 15, and `a, alpha , beta, gamma, b` are in H.P, where `1//alpha + 1//beta + 1//gamma` = 5/3. Find a and b

Options

A. 1 and 9
B. 3 and 7
C. 2 and 8
D. 5 and 6

Answer

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

### Step 1: Understand the given information
We are given that \( a, x, y, z, b \) are in A.P. and \( x + y + z = 15 \). We also know that \( a, \alpha, \beta, \gamma, b \) are in H.P., and \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{5}{3} \).

### Step 2: Express \( x, y, z \) in terms of \( a \) and \( b \)
Since \( a, x, y, z, b \) are in A.P., we can express them as:
- Let \( x = a + d \)
- Let \( y = a + 2d \)
- Let \( z = a + 3d \)
- Let \( b = a + 4d \)

### Step 3: Use the condition \( x + y + z = 15 \)
Substituting the expressions for \( x, y, z \):
\[ (a + d) + (a + 2d) + (a + 3d) = 15 \]
This simplifies to:
\[ 3a + 6d = 15 \]
Dividing the entire equation by 3 gives:
\[ a + 2d = 5 \quad \text{(Equation 1)} \]

### Step 4: Express \( a \) and \( b \) in terms of \( d \)
From Equation 1, we can express \( a \):
\[ a = 5 - 2d \]
And for \( b \):
\[ b = a + 4d = (5 - 2d) + 4d = 5 + 2d \]

### Step 5: Use the H.P. condition
Since \( a, \alpha, \beta, \gamma, b \) are in H.P., it means that \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{a}, \frac{1}{b} \) are in A.P.
Thus, we can write:
\[ \frac{1}{\alpha} + \frac{1}{\gamma} = 2 \cdot \frac{1}{\beta} \]
Also, from the given condition:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{5}{3} \quad \text{(Equation 2)} \]

### Step 6: Substitute values into Equation 2
Using the relationship from A.P.:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{5}{3} \]
We can express \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \) in terms of \( a \) and \( b \):
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{3}{2} \left( \frac{1}{a} + \frac{1}{b} \right) \]
Substituting \( a = 5 - 2d \) and \( b = 5 + 2d \):
\[ \frac{1}{5 - 2d} + \frac{1}{5 + 2d} \]

### Step 7: Solve for \( d \)
Setting up the equation:
\[ \frac{3}{2} \left( \frac{1}{5 - 2d} + \frac{1}{5 + 2d} \right) = \frac{5}{3} \]
Cross-multiplying and simplifying gives:
\[ \frac{3(5 + 2d) + 3(5 - 2d)}{(5 - 2d)(5 + 2d)} = \frac{10}{3} \]
This leads to:
\[ \frac{30}{25 - 4d^2} = \frac{10}{3} \]
Cross-multiplying and solving for \( d \):
\[ 90 = 10(25 - 4d^2) \]
\[ 90 = 250 - 40d^2 \]
\[ 40d^2 = 160 \implies d^2 = 4 \implies d = \pm 2 \]

### Step 8: Find values of \( a \) and \( b \)
1. For \( d = 2 \):
- \( a = 5 - 2(2) = 1 \)
- \( b = 5 + 2(2) = 9 \)

2. For \( d = -2 \):
- \( a = 5 - 2(-2) = 9 \)
- \( b = 5 + 2(-2) = 1 \)

### Conclusion
Thus, the values of \( a \) and \( b \) are:
- \( a = 1, b = 9 \) or \( a = 9, b = 1 \)