A man saved ₹ 33000 in 10 months. In each month after the first, he saved ₹ 100 more than he did in the preceding month. How much did he save in the first month?

Explanation

To solve the problem step by step, we need to determine how much the man saved in the first month given that he saved ₹33,000 over 10 months, increasing his savings by ₹100 each month.

### Step 1: Define the savings in the first month
Let the amount saved in the first month be \( x \) (in rupees).

### Step 2: Define the savings for subsequent months
In the second month, he saves \( x + 100 \).
In the third month, he saves \( x + 200 \).
This pattern continues until the tenth month, where he saves \( x + 900 \).

### Step 3: Write the total savings equation
The total savings over the 10 months can be expressed as:
\[ \text{Total Savings} = x + (x + 100) + (x + 200) + \ldots + (x + 900) \]

### Step 4: Simplify the total savings
We can simplify the total savings:
\[ \text{Total Savings} = 10x + (100 + 200 + 300 + \ldots + 900) \]

### Step 5: Calculate the sum of the arithmetic series
The series \( 100 + 200 + 300 + \ldots + 900 \) is an arithmetic progression (AP) with:
- First term \( a = 100 \)
- Common difference \( d = 100 \)
- Number of terms \( n = 9 \)

The sum \( S_n \) of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Substituting the values:
\[ S_9 = \frac{9}{2} \times (2 \times 100 + (9-1) \times 100) = \frac{9}{2} \times (200 + 800) = \frac{9}{2} \times 1000 = 4500 \]

### Step 6: Substitute the sum back into the total savings equation
Now we can substitute back into the total savings equation:
\[ \text{Total Savings} = 10x + 4500 \]
We know from the problem statement that the total savings is ₹33,000:
\[ 10x + 4500 = 33000 \]

### Step 7: Solve for \( x \)
Subtract 4500 from both sides:
\[ 10x = 33000 - 4500 \]
\[ 10x = 28500 \]
Now, divide by 10:
\[ x = 2850 \]

### Conclusion
Thus, the amount saved in the first month is ₹2850.

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