A sum of Rs. 6240 is paid off in 30 instalments such each instalments is Rs. 10 more than the preceding instalment. The value of the first instalments is

Options

A. Rs. 64
B. Rs. 65
C. Rs. 63
D. Rs. 62

Answer

To solve the problem, we need to find the value of the first installment (let's denote it as \( a \)) given that each subsequent installment is Rs. 10 more than the previous one, and the total sum of all installments is Rs. 6240 over 30 installments.

### Step-by-Step Solution:

1. **Define the first installment**:
Let the first installment be \( a \).

2. **Define the subsequent installments**:
The second installment will be \( a + 10 \), the third installment will be \( a + 20 \), and so on. The \( n^{th} \) installment can be expressed as:
\[ a_n = a + (n-1) \times 10 \]
where \( n \) is the installment number.

3. **Identify the number of installments**:
We have a total of \( n = 30 \) installments.

4. **Use the formula for the sum of an arithmetic series**:
The sum \( S_n \) of the first \( n \) terms of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where:
- \( n \) is the number of terms (30 in this case),
- \( a \) is the first term,
- \( d \) is the common difference (10 in this case).

5. **Substitute the known values into the formula**:
We know the total sum \( S_n = 6240 \), \( n = 30 \), and \( d = 10 \). Plugging these values into the formula gives:
\[ 6240 = \frac{30}{2} \times (2a + (30-1) \times 10) \]

6. **Simplify the equation**:
\[ 6240 = 15 \times (2a + 290) \]
Dividing both sides by 15:
\[ 416 = 2a + 290 \]

7. **Isolate \( 2a \)**:
\[ 2a = 416 - 290 \]
\[ 2a = 126 \]

8. **Solve for \( a \)**:
\[ a = \frac{126}{2} = 63 \]

Thus, the value of the first installment is **Rs. 63**.