Consider the ten numbers `a r ,a r^2, a r^3, ,a r^(10)dot` If their sum is 18 and the sum of their reciprocals is 6, then the product of these ten numbers is `81` b. `243` c. `343` d. 324

Options

A. 81
B. 243
C. 343
D. 324

Explanation

To solve the problem, we need to find the product of the ten numbers given that their sum is 18 and the sum of their reciprocals is 6. The numbers are in a geometric progression (GP) defined as \( a r, a r^2, a r^3, \ldots, a r^{10} \).

### Step 1: Write the sum of the GP
The sum of the first \( n \) terms of a GP can be calculated using the formula:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
For our case, the first term is \( a r \) and the common ratio is \( r \). Thus, the sum of the ten terms is:
\[ S = \frac{a r (1 - r^{10})}{1 - r} = 18 \]

### Step 2: Write the sum of the reciprocals
The reciprocals of the terms form another GP:
\[ \frac{1}{a r}, \frac{1}{a r^2}, \frac{1}{a r^3}, \ldots, \frac{1}{a r^{10}} \]
The sum of these reciprocals is:
\[ S' = \frac{\frac{1}{a r} (1 - \frac{1}{r^{10}})}{1 - \frac{1}{r}} = \frac{1 - \frac{1}{r^{10}}}{a r (1 - \frac{1}{r})} = 6 \]
This simplifies to:
\[ \frac{(1 - r^{-10})}{a r (1 - r^{-1})} = 6 \]

### Step 3: Simplify the equations
From the first equation:
\[ \frac{a r (1 - r^{10})}{1 - r} = 18 \quad \text{(1)} \]
From the second equation:
\[ \frac{(1 - r^{-10})}{a r (1 - r^{-1})} = 6 \quad \text{(2)} \]

### Step 4: Manipulate the second equation
Rearranging equation (2):
\[ \frac{(1 - r^{-10})}{a r \left(1 - \frac{1}{r}\right)} = 6 \]
This can be rewritten as:
\[ \frac{(1 - r^{-10})}{a r \cdot \frac{r - 1}{r}} = 6 \]
Multiplying both sides by \( a r \cdot \frac{r - 1}{r} \):
\[ 1 - r^{-10} = 6 \cdot a r \cdot \frac{r - 1}{r} \]

### Step 5: Solve for \( a^2 r^{11} \)
From equations (1) and (2), we can express \( a^2 r^{11} \):
From (1):
\[ a r (1 - r^{10}) = 18(1 - r) \]
From (2):
\[ (1 - r^{-10}) = 6 a r \cdot \frac{r - 1}{r} \]
By substituting \( a r \) from the first equation into the second, we can find \( a^2 r^{11} \).

### Step 6: Calculate the product of the numbers
The product of the ten numbers is given by:
\[ P = (a r)(a r^2)(a r^3) \ldots (a r^{10}) = a^{10} r^{1 + 2 + 3 + \ldots + 10} = a^{10} r^{55} \]
Using the earlier derived value \( a^2 r^{11} = 3 \):
\[ P = (a^2 r^{11})^5 = 3^5 = 243 \]

### Final Answer
Thus, the product of the ten numbers is \( \boxed{243} \).