Question:
Find the sum of first 15 multiples of 8.
Solution:
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:
Thus, we have:
$l=a+(n-1) d$
$=8+(15-1) 8$
$=120$
$\therefore$ Required sum $=\frac{n}{2}(a+l)$
$=\frac{15}{2}[8+120]=15 \times 64=960$
Hence, the required sum is 960.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.