the terms drop by 11 each time
1994 = 181*11 + 3
So, subtracting eleven 181 more times leaves a remainder of 3
using your usual formula,
S184 = 184/2 (2*2016 + 183(-11)) = 185748
The following series does not go below zero:
2016+2005+1994+...+L
(a) What is the smallest possible value of the last number L ?
(b) Find the sum of the series.
3 answers
Sorry, what is the "usual formula"?
Sn = n/2 (2a+(n-1)d)
Looks like you need to review your AP stuff.
Looks like you need to review your AP stuff.
1 answer