A certain sum under compound interest, in terest being compounded annually earns an interest of₹ 864 and ₹ 1036.80 respectively in the third year and fourth year respectively. Find the sum.

Options

A. ₹ 2400
B. ₹ 3000
C. ₹4000
D. ₹ 3200

Answer

To solve the problem step by step, we will use the information provided about the compound interest earned in the third and fourth years.

### Step 1: Define the Variables
Let:
- \( P \) = Principal amount (the sum we want to find)
- \( R \) = Rate of interest per annum (in percentage)

### Step 2: Write the Formula for Compound Interest
The compound interest for the third year can be calculated as:
\[ \text{CI for 3rd year} = A_3 - A_2 \]
Where:
- \( A_3 \) = Amount at the end of the third year
- \( A_2 \) = Amount at the end of the second year

The amount at the end of \( n \) years is given by:
\[ A_n = P \left(1 + \frac{R}{100}\right)^n \]

Thus, we can express the amounts as:
\[ A_3 = P \left(1 + \frac{R}{100}\right)^3 \]
\[ A_2 = P \left(1 + \frac{R}{100}\right)^2 \]

### Step 3: Set Up the Equation for the Third Year
The compound interest for the third year is given as ₹ 864:
\[ \text{CI for 3rd year} = A_3 - A_2 = P \left(1 + \frac{R}{100}\right)^3 - P \left(1 + \frac{R}{100}\right)^2 = 864 \]

Factoring out \( P \left(1 + \frac{R}{100}\right)^2 \):
\[ P \left(1 + \frac{R}{100}\right)^2 \left(1 + \frac{R}{100} - 1\right) = 864 \]
\[ P \left(1 + \frac{R}{100}\right)^2 \left(\frac{R}{100}\right) = 864 \quad \text{(Equation 1)} \]

### Step 4: Set Up the Equation for the Fourth Year
The compound interest for the fourth year is given as ₹ 1036.80:
\[ \text{CI for 4th year} = A_4 - A_3 = P \left(1 + \frac{R}{100}\right)^4 - P \left(1 + \frac{R}{100}\right)^3 = 1036.80 \]

Factoring out \( P \left(1 + \frac{R}{100}\right)^3 \):
\[ P \left(1 + \frac{R}{100}\right)^3 \left(1 + \frac{R}{100} - 1\right) = 1036.80 \]
\[ P \left(1 + \frac{R}{100}\right)^3 \left(\frac{R}{100}\right) = 1036.80 \quad \text{(Equation 2)} \]

### Step 5: Divide Equation 1 by Equation 2
Now, we can divide Equation 1 by Equation 2:
\[ \frac{P \left(1 + \frac{R}{100}\right)^2 \left(\frac{R}{100}\right)}{P \left(1 + \frac{R}{100}\right)^3 \left(\frac{R}{100}\right)} = \frac{864}{1036.80} \]

This simplifies to:
\[ \frac{1}{1 + \frac{R}{100}} = \frac{864}{1036.80} \]

Calculating the right side:
\[ \frac{864}{1036.80} = \frac{864 \times 100}{103680} = \frac{86400}{103680} = \frac{864}{1036.8} = 0.8333 \]

Thus:
\[ 1 + \frac{R}{100} = \frac{1036.80}{864} \]
Calculating this gives:
\[ 1 + \frac{R}{100} = 1.2 \implies \frac{R}{100} = 0.2 \implies R = 20\% \]

### Step 6: Substitute R Back into Equation 1
Now substituting \( R = 20 \) back into Equation 1:
\[ P \left(1 + \frac{20}{100}\right)^2 \left(\frac{20}{100}\right) = 864 \]
\[ P \left(1.2\right)^2 \left(0.2\right) = 864 \]
\[ P \cdot 1.44 \cdot 0.2 = 864 \]
\[ P \cdot 0.288 = 864 \]
\[ P = \frac{864}{0.288} = 3000 \]

### Final Answer
The sum (principal amount) is ₹ 3000.

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