Find the sum of the first 15 multiples of 8.

First 15 multiples of 8 are 8, 16, 24, …, 120.

Sum of these numbers forms an arithmetic series 8 + 16 + 24 + … + 120.

Here, first term = a = 8

Common difference = d = 8

Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 15 terms of this arithmetic series is given by:

∴ S₁₅ = [2(8) + (15 - 1)(8)]

= (15/2) [16 + 112]

=(15/2) × 128

= 15 × 64

= 960