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:
Sn = [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