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Old 10-29-2011   #3
MAS1

 
Join Date: Dec 2008
Posts: 249
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Quote:
Originally Posted by Pi= View Post
Alright, I have homework I need finished by tommorrow. Here is a problem or two.
7. Expressed in terms of n what is the sum of all the terms in the arithmetic sequence (n-7), (n-2), . . . . (n+428)?
8. If x = a/b, find (a+b)/(a-b) in terms of x.
7. If the first term is n - 7 and the second term is n - 2, then I am assuming the common difference, d, is +5.

The sum, Sn, of an arithmetic progression is equal to (n/2)(a1 + an) where a1 is the 1st term and an is the nth term. So we need to find n.

an = a1 + d(n - 1)
Substituting gives:
(n + 428) = (n - 7) + 5(n - 1)
n + 428 = n - 7 + 5n - 5
n + 428 = 6n - 12
440 = 5n
n = 88

So n + 428 is the 88th term in the progression.

Then the sum of the 1st 88 terms in the progression, S88, is given by:

S88 = (88/2)(n - 7 + n + 428)
S88 = 44(2n + 421)
S88 = 88n + 18524

Hope this helps.
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