A man saves Rs. 3200 during the first year, Rs. 3600 in the next year Rs. 4000 in the third year. If he continues his saving in this sequence, in how many years will he have Rs. 200000?

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the pattern of savings
The man saves:
- Rs. 3200 in the first year
- Rs. 3600 in the second year
- Rs. 4000 in the third year

The difference between the savings in consecutive years is:
- 3600 - 3200 = 400
- 4000 - 3600 = 400

This shows that the savings form an arithmetic progression (AP) with:
- First term \( a = 3200 \)
- Common difference \( d = 400 \)

### Step 2: Write the formula for the sum of the first \( n \) terms of an AP
The sum \( S_n \) of the first \( n \) terms of an arithmetic progression is given by the formula:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \]

### Step 3: Set up the equation for total savings
We need to find \( n \) such that the total savings equals Rs. 200,000:
\[ S_n = 200000 \]
Substituting the values of \( a \) and \( d \):
\[ \frac{n}{2} \left( 2 \cdot 3200 + (n-1) \cdot 400 \right) = 200000 \]

### Step 4: Simplify the equation
First, simplify the equation:
\[ \frac{n}{2} \left( 6400 + 400(n-1) \right) = 200000 \]
\[ \frac{n}{2} \left( 6400 + 400n - 400 \right) = 200000 \]
\[ \frac{n}{2} \left( 6000 + 400n \right) = 200000 \]
Multiply both sides by 2 to eliminate the fraction:
\[ n(6000 + 400n) = 400000 \]
\[ 6000n + 400n^2 = 400000 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 400n^2 + 6000n - 400000 = 0 \]
Dividing the entire equation by 100:
\[ 4n^2 + 60n - 4000 = 0 \]

### Step 6: Use the quadratic formula
We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = 60 \), and \( c = -4000 \):
\[ n = \frac{-60 \pm \sqrt{60^2 - 4 \cdot 4 \cdot (-4000)}}{2 \cdot 4} \]
\[ n = \frac{-60 \pm \sqrt{3600 + 64000}}{8} \]
\[ n = \frac{-60 \pm \sqrt{67600}}{8} \]
\[ n = \frac{-60 \pm 260}{8} \]

### Step 7: Calculate the possible values for \( n \)
Calculating the two possible values:
1. \( n = \frac{200}{8} = 25 \)
2. \( n = \frac{-320}{8} = -40 \) (not valid since \( n \) cannot be negative)

### Final Answer
Thus, the number of years it will take for the man to save Rs. 200,000 is:
\[ \boxed{25} \]