After paying 30 out of 40 installments of a debt of Rs. 3600, one third of the debt is unpaid. If the installments are forming an arithmetic series, then what is the first instalment ?

Options

A. Rs. 50
B. Rs. 51
C. Rs. 105
D. Rs. 110

Explanation

To solve the problem, we need to find the first installment of an arithmetic series given that after paying 30 installments of a total debt of Rs. 3600, one-third of the debt remains unpaid.

### Step-by-Step Solution:

1. **Understanding the Debt and Installments**:
- Total debt = Rs. 3600
- One-third of the debt is unpaid, which means:
\[ \text{Unpaid Debt} = \frac{1}{3} \times 3600 = 1200 \]
- Therefore, the amount paid after 30 installments is:
\[ \text{Amount Paid} = 3600 - 1200 = 2400 \]

2. **Finding the Total Number of Installments**:
- There are 40 installments in total.

3. **Arithmetic Series Formula**:
- The installments form an arithmetic series where:
- First installment = \( A \)
- Common difference = \( d \)
- The \( n \)-th term of an arithmetic series can be expressed as:
\[ a_n = A + (n-1)d \]

4. **Calculating the Sum of the First 30 Installments**:
- The sum \( S_n \) of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \times (2A + (n-1)d) \]
- For the first 30 installments:
\[ S_{30} = \frac{30}{2} \times (2A + 29d) = 15 \times (2A + 29d) \]
- We know that this sum equals Rs. 2400:
\[ 15 \times (2A + 29d) = 2400 \]
- Dividing both sides by 15:
\[ 2A + 29d = 160 \quad \text{(Equation 1)} \]

5. **Calculating the Total Sum of All 40 Installments**:
- The total sum of all 40 installments:
\[ S_{40} = \frac{40}{2} \times (2A + 39d) = 20 \times (2A + 39d) \]
- This total sum equals Rs. 3600:
\[ 20 \times (2A + 39d) = 3600 \]
- Dividing both sides by 20:
\[ 2A + 39d = 180 \quad \text{(Equation 2)} \]

6. **Solving the Equations**:
- Now we have two equations:
- Equation 1: \( 2A + 29d = 160 \)
- Equation 2: \( 2A + 39d = 180 \)
- Subtract Equation 1 from Equation 2:
\[ (2A + 39d) - (2A + 29d) = 180 - 160 \]
\[ 10d = 20 \]
\[ d = 2 \]

7. **Substituting Back to Find \( A \)**:
- Substitute \( d = 2 \) back into Equation 1:
\[ 2A + 29(2) = 160 \]
\[ 2A + 58 = 160 \]
\[ 2A = 160 - 58 \]
\[ 2A = 102 \]
\[ A = 51 \]

### Final Answer:
The first installment \( A \) is Rs. 51.