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 installment ?

Options

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

Answer

To find the first installment of a debt of Rs. 3600 paid in 40 installments forming an arithmetic series, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Total Debt and Unpaid Amount**:
The total debt is Rs. 3600. According to the problem, one third of the debt is unpaid after 30 installments.
\[ \text{Unpaid Debt} = \frac{1}{3} \times 3600 = 1200 \]
Therefore, the amount paid after 30 installments is:
\[ \text{Paid Amount} = 3600 - 1200 = 2400 \]

2. **Use the Formula for the Sum of an Arithmetic Series**:
The sum of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \times (a + l) \]
where \( a \) is the first term, \( l \) is the last term, and \( n \) is the number of terms. For the first 30 installments:
\[ S_{30} = 2400 \]
Substituting \( n = 30 \):
\[ 2400 = \frac{30}{2} \times (a + l_{30}) \quad \text{(Equation 1)} \]
Simplifying gives:
\[ 2400 = 15 \times (a + l_{30}) \implies a + l_{30} = \frac{2400}{15} = 160 \quad \text{(Equation 1)} \]

3. **Calculate the Total Sum for 40 Installments**:
For the total 40 installments:
\[ S_{40} = 3600 \]
Substituting \( n = 40 \):
\[ 3600 = \frac{40}{2} \times (a + l_{40}) \quad \text{(Equation 2)} \]
Simplifying gives:
\[ 3600 = 20 \times (a + l_{40}) \implies a + l_{40} = \frac{3600}{20} = 180 \quad \text{(Equation 2)} \]

4. **Set Up the Equations**:
Now we have two equations:
- Equation 1: \( a + l_{30} = 160 \)
- Equation 2: \( a + l_{40} = 180 \)

5. **Subtract the Two Equations**:
Subtract Equation 1 from Equation 2:
\[ (a + l_{40}) - (a + l_{30}) = 180 - 160 \]
This simplifies to:
\[ l_{40} - l_{30} = 20 \]

6. **Find the Relationship Between \( l_{30} \) and \( l_{40} \)**:
The last term \( l_{n} \) of an arithmetic series can be expressed as:
\[ l_n = a + (n-1)d \]
Therefore:
\[ l_{30} = a + 29d \quad \text{and} \quad l_{40} = a + 39d \]
Substituting these into the difference:
\[ (a + 39d) - (a + 29d) = 20 \implies 10d = 20 \implies d = 2 \]

7. **Substitute \( d \) Back to Find \( a \)**:
Now substitute \( d = 2 \) back into either equation. Using Equation 1:
\[ a + (a + 29 \cdot 2) = 160 \]
Simplifying gives:
\[ a + a + 58 = 160 \implies 2a + 58 = 160 \implies 2a = 102 \implies a = 51 \]

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