A person is entitled to receive an annual payment which for each year is less by one tenth of what it was for the year before . If the first first payment is 100, then find the maximum possible payment which be can receive however long he may live :

Options

A. 900
B. 9999
C. 1000
D. none of these

Answer

To solve the problem, we need to determine the total payments a person receives over the years, given that each subsequent payment is reduced by one-tenth of the previous payment. Let's break down the solution step by step.

### Step 1: Identify the first payment
The first payment is given as Rs. 100.

### Step 2: Calculate the second payment
The second payment is reduced by one-tenth of the first payment. Therefore, the second payment can be calculated as:
\[ \text{Second Payment} = \text{First Payment} - \left(\frac{1}{10} \times \text{First Payment}\right) = 100 - 10 = 90 \]

### Step 3: Calculate the third payment
The third payment is also reduced by one-tenth of the second payment:
\[ \text{Third Payment} = \text{Second Payment} - \left(\frac{1}{10} \times \text{Second Payment}\right) = 90 - 9 = 81 \]

### Step 4: Calculate the fourth payment
Continuing this pattern, the fourth payment is:
\[ \text{Fourth Payment} = \text{Third Payment} - \left(\frac{1}{10} \times \text{Third Payment}\right) = 81 - 8.1 = 72.9 \]

### Step 5: Identify the pattern
From the calculations, we can see that each payment is \( \frac{9}{10} \) of the previous payment. This forms a geometric series where:
- The first term \( a = 100 \)
- The common ratio \( r = \frac{9}{10} \)

### Step 6: Sum the infinite geometric series
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{100}{1 - \frac{9}{10}} = \frac{100}{\frac{1}{10}} = 1000 \]

### Conclusion
The maximum possible payment that the person can receive, however long he may live, is Rs. 1000.