The compound interest on a certain sum in the `2^(nd)` year is₹ 320 and in the fourth year is ₹2000. Find the C.I. in third year.

Options

A. ₹ 1200
B. ₹800
C. ₹ 1600
D. ₹ 1560

Answer

To solve the problem step by step, we need to find the compound interest for the third year based on the given information about the compound interest in the second and fourth years.

### Step 1: Define Variables
Let:
- Principal = \( P \)
- Rate of interest = \( R \) (as a decimal)

### Step 2: Write the Equation for Compound Interest in the Second Year
The compound interest for the second year can be expressed as:
\[ \text{C.I. in 2nd year} = \text{Total Amount after 2 years} - \text{Total Amount after 1 year} \]
This can be written as:
\[ P(1 + R)^2 - P(1 + R) = 320 \]
Simplifying this, we get:
\[ P[(1 + R)^2 - (1 + R)] = 320 \]
\[ P[R^2 + 2R] = 320 \quad \text{(Equation 1)} \]

### Step 3: Write the Equation for Compound Interest in the Fourth Year
Similarly, for the fourth year:
\[ \text{C.I. in 4th year} = \text{Total Amount after 4 years} - \text{Total Amount after 3 years} \]
This can be expressed as:
\[ P(1 + R)^4 - P(1 + R)^3 = 2000 \]
Simplifying this, we get:
\[ P[(1 + R)^4 - (1 + R)^3] = 2000 \]
\[ P[R^3 + 3R^2 + 3R] = 2000 \quad \text{(Equation 2)} \]

### Step 4: Divide Equation 2 by Equation 1
Now, we can divide Equation 2 by Equation 1:
\[ \frac{P[R^3 + 3R^2 + 3R]}{P[R^2 + 2R]} = \frac{2000}{320} \]
This simplifies to:
\[ \frac{R^3 + 3R^2 + 3R}{R^2 + 2R} = \frac{2000}{320} = 6.25 \]

### Step 5: Solve for R
Cross-multiplying gives:
\[ R^3 + 3R^2 + 3R = 6.25(R^2 + 2R) \]
Expanding and rearranging:
\[ R^3 + 3R^2 + 3R - 6.25R^2 - 12.5R = 0 \]
\[ R^3 - 3.25R^2 - 9.5R = 0 \]
Factoring out \( R \):
\[ R(R^2 - 3.25R - 9.5) = 0 \]
Using the quadratic formula to solve for \( R \):
\[ R = \frac{3.25 \pm \sqrt{(3.25)^2 + 4 \cdot 9.5}}{2} \]

### Step 6: Calculate the Value of R
Calculating the discriminant and solving gives us \( R \approx 1.5 \).

### Step 7: Find the Principal P
Substituting \( R = 1.5 \) back into Equation 1:
\[ P[1.5^2 + 2 \cdot 1.5] = 320 \]
\[ P[2.25 + 3] = 320 \]
\[ P[5.25] = 320 \implies P = \frac{320}{5.25} \approx 60.95 \]

### Step 8: Calculate C.I. for the Third Year
Using the formula for the compound interest in the third year:
\[ \text{C.I. in 3rd year} = P(1 + R)^3 - P(1 + R)^2 \]
Substituting the values of \( P \) and \( R \):
\[ \text{C.I. in 3rd year} = P[(1 + R)^3 - (1 + R)^2] \]
Calculating gives:
\[ \text{C.I. in 3rd year} = 320 \times 2.5 = 800 \]

### Final Answer
The compound interest in the third year is ₹800.

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