x+y+z=15 if 9, x, y, z, a are in A.P. while `(1)/(X)+(1)/(Y)+(1)/(Z)=(5)/(3)if9,X, Y, Z, a` are in H.P., then the value of a will be -

Options

A. 1
B. 2
C. 3
D. 9

Explanation

To solve the problem step by step, we will use the given conditions that \( x + y + z = 15 \) and the sequences involving \( 9, x, y, z, a \) being in Arithmetic Progression (A.P.) and \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3} \) with \( 9, x, y, z, a \) being in Harmonic Progression (H.P.).

### Step 1: Express \( x + y + z \) in terms of \( a \)
Given that \( 9, x, y, z, a \) are in A.P., we can express \( x, y, z \) in terms of \( a \).

The formula for the sum of terms in an A.P. is:
\[ x + y + z = \frac{3}{2}(9 + a) \]

### Step 2: Set up the equation
From the problem, we know:
\[ x + y + z = 15 \]
Thus, we can set up the equation:
\[ \frac{3}{2}(9 + a) = 15 \]

### Step 3: Solve for \( a \)
To solve for \( a \), multiply both sides by \( 2 \):
\[ 3(9 + a) = 30 \]
Now, divide by \( 3 \):
\[ 9 + a = 10 \]
Subtract \( 9 \) from both sides:
\[ a = 1 \]

### Step 4: Verify with the Harmonic Progression condition
Now we need to check the second condition that \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3} \) when \( 9, x, y, z, a \) are in H.P.

Since \( 9, x, y, z, a \) are in H.P., it implies that \( \frac{1}{9}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{a} \) are in A.P.

Using the A.P. property:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{2} \left( \frac{1}{9} + \frac{1}{a} \right) \]
Substituting \( a = 1 \):
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{2} \left( \frac{1}{9} + 1 \right) = \frac{3}{2} \left( \frac{1 + 9}{9} \right) = \frac{3}{2} \left( \frac{10}{9} \right) = \frac{15}{9} = \frac{5}{3} \]

### Conclusion
Both conditions are satisfied, confirming that the value of \( a \) is indeed:
\[ \boxed{1} \]