1 Answers
The first 15 multiples of 8 are 8, 16, 24, 32,…….
This is an AP in which a = 8, d = (16 - 8) = 8 and n = 15.
Thus, we have:
l = a + (n - 1)d
= 8 + (15 - 1)8
= 120
∴ Required sum = \(\frac{n}{2}(a+l)\)
= \(\frac{15}{2}[8+120]=15\times64=960\)
Hence, the required sum is 960.
4 views
Answered