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

x-4 + 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+(n-1)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) - (n-3) = 75

substituting the formula,

(n/2)(2a+(n-1)d) - [(n-3)/2][2a+(n-3-1)d] = 75

solve and find n.

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