Find the sum of the first 15 multiples of 8.
\(960\)
Step 1: Recognise the AP.
The multiples of 8 form an arithmetic progression:
\[8, 16, 24, 32, \ldots\]
Here, the first term \(a = 8\) and the common difference \(d = 8\).
We need the sum of the first 15 terms, i.e., \(S_{15}\).
Step 2: Use the formula for the sum of the first n terms of an AP.
\[S_n = \dfrac{n}{2}[2a + (n - 1)d]\]
Substitute \(n = 15\), \(a = 8\), and \(d = 8\):
\[S_{15} = \dfrac{15}{2}[2 \cdot 8 + 14 \cdot 8]\]
Step 3: Simplify inside the brackets.
\[2a = 16\]
\[(n - 1)d = 14 \cdot 8 = 112\]
So:
\[S_{15} = \dfrac{15}{2}(16 + 112) = \dfrac{15}{2} \cdot 128\]
Step 4: Final computation.
\[S_{15} = 15 \cdot 64 = 960\]
Conclusion: The sum of the first 15 multiples of 8 is 960.