Arithmetic Series: 25 on 1st, 50 on 2nd, 75 on 3rd, etc.  TOPIC_SOLVED

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Arithmetic Series: 25 on 1st, 50 on 2nd, 75 on 3rd, etc.

Postby tecnikal on Sat Mar 28, 2009 1:52 am

Im taking up advanced functions and relations and this questions (for some reason) has me stumped. I think i got the right answer but the answer in the back of my book is different:

Suppose you earn 25 cents on September 1, 50 cents on September 2, 75 cents on September 3, and so on. How much money will you earn by the end of the month?

The answer in the back is 116.25$.

I got 120.

Btw this has to deal with a general formula:

t(n)= a + (n-1)d where A is the first term and D is the difference between each term.
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Postby stapel_eliz on Sat Mar 28, 2009 2:51 pm

tecnikal wrote:Suppose you earn 25 cents on September 1, 50 cents on September 2, 75 cents on September 3, and so on. How much money will you earn by the end of the month?

The answer in the back is 116.25$. I got 120.

Unfortunately, it's kinda hard to find errors in work that we can't see.... :oops:

The formula you listed is for the value of the n-th term t(n) of the arithmetic sequence with first term a1 = a (assuming, contrary to standard practice, that you mean "A" and "a" to mean the same thing) and common difference d. Then the last term (since September has thirty days) is:

. . . . .

. . . . .

Now you need to apply the formula for the sum s(n) of the first n terms of the arithmetic series:

. . . . .

Plug the values of a1, a[sub]n[sub], and n into the formula:

. . . . .

. . . . . . . . . .

Simplify and compare with the book's answer. :wink:
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