A man arranges to pay off a debt of ` 3,600 in 40 annual instalments which form an AP. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first instalment.

Options

A. 55
B. 51
C. 53
D. 49

Answer

To solve the problem step by step, we will use the properties of arithmetic progressions (AP) and the formula for the sum of the first n terms of an AP.

### Step 1: Understand the total debt and the number of installments
The total debt is `3600`, and it is to be paid off in `40` annual installments.

### Step 2: Set up the equation for the total sum of installments
The sum of the first `n` terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Where:
- \( S_n \) is the sum of the first `n` terms,
- \( a \) is the first term (the first installment),
- \( d \) is the common difference,
- \( n \) is the number of terms.

For our case:
\[ S_{40} = 3600 \]
Substituting into the formula:
\[ 3600 = \frac{40}{2} \times (2a + 39d) \]
This simplifies to:
\[ 3600 = 20 \times (2a + 39d) \]
Thus, we have:
\[ 2a + 39d = 180 \quad \text{(Equation 1)} \]

### Step 3: Analyze the situation after 30 installments
After 30 installments, one-third of the debt is unpaid. Therefore, the amount paid after 30 installments is:
\[ \text{Unpaid amount} = \frac{1}{3} \times 3600 = 1200 \]
Thus, the amount paid is:
\[ \text{Paid amount} = 3600 - 1200 = 2400 \]
Using the sum formula for the first 30 installments:
\[ S_{30} = 2400 \]
Substituting into the formula:
\[ 2400 = \frac{30}{2} \times (2a + 29d) \]
This simplifies to:
\[ 2400 = 15 \times (2a + 29d) \]
Thus, we have:
\[ 2a + 29d = 160 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( 2a + 39d = 180 \) (Equation 1)
2. \( 2a + 29d = 160 \) (Equation 2)

Subtract Equation 2 from Equation 1:
\[ (2a + 39d) - (2a + 29d) = 180 - 160 \]
This simplifies to:
\[ 10d = 20 \]
Thus, we find:
\[ d = 2 \]

### Step 5: Substitute \( d \) back to find \( a \)
Now substitute \( d = 2 \) back into Equation 2:
\[ 2a + 29(2) = 160 \]
This simplifies to:
\[ 2a + 58 = 160 \]
Thus:
\[ 2a = 160 - 58 \]
\[ 2a = 102 \]
\[ a = 51 \]

### Conclusion
The value of the first installment is \( \text{Rs. } 51 \).

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