Sum of first 15 multiples of 8 is

Options

A. 960
B. 660
C. 1200
D. 1060

Answer

To find the sum of the first 15 multiples of 8, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the First 15 Multiples of 8**:
The first 15 multiples of 8 can be expressed as:
- 1st multiple: \( 8 \times 1 = 8 \)
- 2nd multiple: \( 8 \times 2 = 16 \)
- 3rd multiple: \( 8 \times 3 = 24 \)
- ...
- 15th multiple: \( 8 \times 15 = 120 \)

So, the first 15 multiples of 8 are: \( 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120 \).

2. **Recognize the Pattern**:
These multiples form an arithmetic progression (AP) where:
- First term \( a = 8 \)
- Common difference \( d = 8 \)
- Number of terms \( n = 15 \)

3. **Use the Formula for the Sum of an AP**:
The sum \( S_n \) of the first \( n \) terms of an arithmetic progression can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (a + l) \]
where \( l \) is the last term.

4. **Calculate the Last Term**:
The last term \( l \) (15th multiple of 8) is:
\[ l = 8 \times 15 = 120 \]

5. **Substitute the Values into the Formula**:
Now substitute \( n = 15 \), \( a = 8 \), and \( l = 120 \) into the sum formula:
\[ S_{15} = \frac{15}{2} \times (8 + 120) \]

6. **Simplify the Expression**:
Calculate \( 8 + 120 = 128 \):
\[ S_{15} = \frac{15}{2} \times 128 \]

7. **Perform the Multiplication**:
Now calculate \( \frac{15 \times 128}{2} \):
\[ S_{15} = \frac{15 \times 128}{2} = 15 \times 64 \]

8. **Final Calculation**:
Calculate \( 15 \times 64 \):
\[ S_{15} = 960 \]

Thus, the sum of the first 15 multiples of 8 is **960**.