The sum of an infinite GP is 162 and the sum of its first n terms is 160. If the inverse of its common ratio is an integer, then how many values of common ratio is/are possible, common ratio is greater than 0?

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to analyze the given information about the infinite geometric progression (GP).

### Step 1: Understand the formulas
The sum of an infinite GP is given by the formula:
\[ S_{\infty} = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

The sum of the first \( n \) terms of a GP is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

### Step 2: Set up the equations
From the problem, we know:
1. \( S_{\infty} = 162 \)
2. \( S_n = 160 \)

Using the formula for the sum of an infinite GP:
\[ \frac{a}{1 - r} = 162 \quad \text{(1)} \]

Using the formula for the sum of the first \( n \) terms:
\[ \frac{a(1 - r^n)}{1 - r} = 160 \quad \text{(2)} \]

### Step 3: Express \( a \) from equation (1)
From equation (1):
\[ a = 162(1 - r) \quad \text{(3)} \]

### Step 4: Substitute \( a \) in equation (2)
Now substitute equation (3) into equation (2):
\[ \frac{162(1 - r)(1 - r^n)}{1 - r} = 160 \]
This simplifies to:
\[ 162(1 - r^n) = 160 \]

### Step 5: Solve for \( r^n \)
Rearranging gives:
\[ 1 - r^n = \frac{160}{162} \]
\[ 1 - r^n = \frac{80}{81} \]
Thus:
\[ r^n = 1 - \frac{80}{81} = \frac{1}{81} \quad \text{(4)} \]

### Step 6: Analyze equation (4)
From equation (4), we have:
\[ r^n = \left(\frac{1}{3}\right)^4 \]
This implies:
\[ r = \frac{1}{3} \quad \text{(for } n = 4\text{)} \]

### Step 7: Consider the inverse of \( r \)
The problem states that the inverse of the common ratio \( r \) is an integer:
\[ \frac{1}{r} = 3 \quad \text{(which is an integer)} \]

### Step 8: Check for other values
Since \( r^n = \frac{1}{81} \), we can also express it as:
\[ r^n = \left(\frac{1}{3}\right)^4 \]
This means \( r \) could also be:
- \( r = \frac{1}{9} \) (for \( n = 2 \))
- \( r = \frac{1}{27} \) (for \( n = 3 \))

### Step 9: List possible values of \( r \)
The possible values of \( r \) that satisfy the conditions are:
1. \( r = \frac{1}{3} \)
2. \( r = \frac{1}{9} \)
3. \( r = \frac{1}{27} \)

### Conclusion
Thus, the total number of possible values for the common ratio \( r \) that are greater than 0 and whose inverse is an integer is **3**.