Posted by Watermelon on Thursday, November 24, 2011 at 7:04am.
Given an arithmetic progression 7,3,1,..., state three consecutive terms in this progression which sum up to 75.

AP calculus  Damon, Thursday, November 24, 2011 at 7:36am
+ 4 = change from term to term
x4 + x + x+4 = 75
3 x = 75
x = 25
so
21, 25, 29

AP calculus  Andrew, Thursday, November 24, 2011 at 7:39am
Think about how you can locate these three consecutive terms.
What is the first term, and the common difference?
The sum of an AP for the first n terms is given by
S(n) = (n/2)(2a+(n1)d)
where a = 7, d = 4
So lets say the 3 consecutive terms are the 14,15 and 16th term.
to get the sum of 75 of these 2 terms, i take
S(16)  S(13)
hence,
S(n)  (n3) = 75
substituting the formula,
(n/2)(2a+(n1)d)  [(n3)/2][2a+(n31)d] = 75
solve and find n.
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