An arithmetic series is the sum of the terms of an arithmetic sequence. A geometric series is the sum of the terms of a geometric sequence. There are other types of series, but you're unlikely to work with them much until you're in calculus. For now, you'll probably mostly work with these two. This page explains and illustrates how to work with arithmetic series.
For reasons that will be explained in calculus, you can only take the "partial" sum of an arithmetic sequence. The partial sum is the sum of a limited (that is to say, a finite) number of terms, like the first ten terms, or the fifth through the hundredth terms.
The formula for the first n terms of an arithmetic sequence, starting with i = 1, is:
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If you take the "2" on the right-hand side of the "equals" sign from under the n and convert it to being a one-half multiplied on the parentheses, you can see that the formula for the sum is, in effect, n times the "average" of the first and last terms.
Thinking of the summation formula this way can be a useful way of memorizing the formula. (By the way: The summation formula can be proved using induction.)
The sum of the first n terms of a series is called "the n-th partial sum", and is often denoted as "S_{n}".
The 35th partial sum of this sequence is the sum of the first thirty-five terms. The first few terms of the sequence are:
The terms have a common difference , so this is indeed an arithmetic sequence. The last term in the partial sum will be:
Then, plugging into the formula, the 35th partial sum is:
Then my answer is:
35th partial sum: S35 = 350
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I could have found the common difference in the above sequence simply by looking at the formula for the sequence's terms. Because this is an arithmetic sequence, then each term is a fixed amount larger than the previous term. If we'd been using a continuous variable, such as the "x" we used when graphing straight lines, instead of the discrete variable n, then " " would be a straight line that increased by one-half at each step.
We can use what we learned about the slope of a straight line, and how this relates to the equation for a straight line, to read off the common difference from the formula for the terms... which can save some time on the test.
From the formula, "2n – 5", for the n-th term, I can see that each term will be two units larger than the previous term. (If I weren't sure about this, I could always plug in some values for n to confirm.) So this is indeed an arithmetical sum. But this summation starts at n = 15, not at n = 1, and the summation formula applies to sums starting at n = 1. So how can I work with this summation? By using a little trick:
The quickest way to find the value of this sum is to find the 14th and 47th partial sums, and then subtract the 14th from the 47th. S14 is the sum of the first through the fourteenth terms. By doing this subtraction, I'll have deducted the first through fourteenth terms from the first through forty-seventh, so I'll be left with the sum of the 15th through 47th terms.
The first term is:
a1 = 2(1) – 5 = –3
The other necessary terms are the fourteenth and the forty-seventh:
a14 = 2(14) – 5 = 23
a47 = 2(47) – 5 = 89
With these values, I now have everything I need in order to find the two partial sums for my subtraction:
Subtracting, I get:
Then my answer is:
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By the way, this subtraction could also be expressed as "S_{47} – S_{14}". Don't be surprised if you see an exercise which uses this notation and expects you to extract the meaning of it before you can begin your computations.
Formatting note: Since this was just a summation, it's safe to assume that "2n – 5" is the expression being summed. However (and especially if you're dealing with something more complex), sometimes grouping symbols may be necessary to make the meaning clear. Properly, the author of the previous exercise should have formatted the summation with grouping symbols, as follows:
I know that the first term has the value:
a1 = 0.25(1) + 2 = 2.25
I can see from the formula that each term will be 0.25 units bigger than the previous term, so this is an arithmetical series, with d = 0.25. Then the summation formula for arithmetical series gives me:
n (2.25 + 0.25n + 2) = 42
n (0.25n + 4.25) = 42
0.25n^{2} + 4.25n – 42 = 0
n^{2} + 17n – 168 = 0
(n + 24)(n – 7) = 0
Solving this quadratic equation, I get that n = –24 (which makes no sense in this context) or n = 7. Then:
n = 7
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You could do the above exercise by adding terms until you get to the required total of 21. But your instructor could easily give you a summation that requires, say, eighty-six terms, or a thousand, before you get to the right total. Even if you could do all the steps, it would be a ridiculous waste of time, especially on a test. So make sure you can do the computations from the formula.
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Checking the terms, I can see that this is indeed an arithmetic series:
5 – 1 = 4
9 – 5 = 4
53 – 49 = 4
(And I want to get in the habit of checking like this, because they won't always tell me, especially on the test, which kind of series they've given me.)
They've given me the first and the last terms of this series, but how many terms are there in total? I'll have to figure this out for myself.
I have the formula for the n-th term of an arithmetic sequence:
a_{n} = a1 + (n – 1)d
They've given me a1 = 1 and I've figured out that d = 4. Plugging these values into the formula, I can figure out how many terms there are:
a_{n} = a1 + (n – 1)d
53 = 1 + (n – 1)(4)
53 = 1 + 4n – 4
53 = 4n – 3
56 = 4n
14 = n
So there are 14 terms in this series. Now I have all the information I need to find the sum, even though I don't know all the terms that are being summed:
1 + 5 + 9 + ... + 49 + 53
Then my answer is:
partial sum S14 = 378
Next, we consider geometric series.
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