Step-by-Step Solution
Key Concept: The given numbers form an arithmetic progression (AP) with first term $a = 8$ and common difference $d = 8$. The sum of $n$ terms of an AP is given by $S_n = \frac{n}{2}\bigl[2a+(n-1)d\bigr]$ or equivalently $S_n = \frac{n}{2}(a + l)$ where $l$ is the last term.
1. Identify the AP:
- First term $a = 8$
- Common difference $d = 8$
- Number of terms $n = 15$
2. Find the last term $l$ (the 15th multiple of 8):
$$l = a + (n-1)d = 8 + (15-1)\times 8 = 8 + 14\times 8 = 8 + 112 = 120$$
3. Use the sum formula for an AP:
$$S_{15} = \frac{n}{2}\,(a + l) = \frac{15}{2}\,(8 + 120)$$
4. Compute the value:
$$S_{15} = \frac{15}{2}\times 128 = 15 \times 64 = 960$$
Thus, the sum of the first 15 multiples of 8 is $960$.
Correct Answer: 960