A sum of Rs 12500 is deposited for `1 (1)/(2)` years, compounded half-yearly. It amounts to 13000 at the end of first half-year. Find :
the final amount. Give your answer correct to the nearest rupee.

Answer

To solve the problem step by step, we will follow the process of calculating the compound interest and the final amount.

### Step 1: Identify the given values
- Principal (P) = Rs 12,500
- Amount after the first half-year (A1) = Rs 13,000
- Total time (T) = 1.5 years = 3/2 years
- Since it is compounded half-yearly, we will consider the time in half-year intervals. Thus, T = 3 half-years.

### Step 2: Find the rate of interest (r)
We know that the amount after the first half-year is given by the formula:
\[ A = P \left(1 + \frac{r}{200}\right) \]
Where:
- A = Amount after the first half-year = Rs 13,000
- P = Principal = Rs 12,500

Substituting the values into the formula:
\[ 13000 = 12500 \left(1 + \frac{r}{200}\right) \]

### Step 3: Solve for r
Rearranging the equation:
\[ 1 + \frac{r}{200} = \frac{13000}{12500} \]
\[ 1 + \frac{r}{200} = 1.04 \]

Now, subtract 1 from both sides:
\[ \frac{r}{200} = 0.04 \]

Multiply both sides by 200 to find r:
\[ r = 0.04 \times 200 = 8 \]

Thus, the rate of interest (r) is 8%.

### Step 4: Calculate the final amount after 1.5 years
Now we will calculate the final amount (A) after 1.5 years using the formula:
\[ A = P \left(1 + \frac{r}{200}\right)^{2t} \]

Where:
- P = Rs 12,500
- r = 8%
- t = 1.5 years

Substituting the values:
\[ A = 12500 \left(1 + \frac{8}{200}\right)^{2 \times \frac{3}{2}} \]
\[ A = 12500 \left(1 + 0.04\right)^{3} \]
\[ A = 12500 \left(1.04\right)^{3} \]

### Step 5: Calculate \( (1.04)^3 \)
Calculating \( (1.04)^3 \):
\[ (1.04)^3 = 1.124864 \]

### Step 6: Calculate the final amount A
Now substituting back to find A:
\[ A = 12500 \times 1.124864 \]
\[ A = 14060.8 \]

### Step 7: Round to the nearest rupee
The final amount rounded to the nearest rupee is:
\[ A \approx 14061 \]

### Final Answer
The final amount after 1.5 years is Rs 14,061.

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