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December 21, 2014

December 21, 2014

Posted by **vincecarl** on Saturday, November 5, 2011 at 2:52am.

- math -
**vincecarl**, Saturday, November 5, 2011 at 2:55amif 5 arithmetic means are inserted between 7 and 25 what is the middle mean to be inserted

- math -
**Reiny**, Saturday, November 5, 2011 at 6:10amThe 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+(n-1)d

99 = 3 + (n-1)(3)

96 = 3n-3

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

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