math
posted by vincecarl on .
find the sum of the integers between 2 and 100 which are divisible by 3

if 5 arithmetic means are inserted between 7 and 25 what is the middle mean to be inserted

The multiples of 3 between 2 and 100 are
3, 6, 9, ... , 99 , an AS where a=3 and d=3
how many of those are there?
t(n) = a+(n1)d
99 = 3 + (n1)(3)
96 = 3n3
99 = 3n
n = 33
so now you want the sum of those 33 arithmetic terms
S(33) = (33/2)(first + last) = (33/2)(3+99) = 1683
Your second question....
so your 7 becomes the first term, and your 25 becomes the 7th term
25 = 7 + 6d
d = 3
so your middle term would be term(4)
= a+3d = 7+9 = 16
check: 7 10 13 16 19 22 25 
1584