Arithmetic Series (page 4 of 5)
An arithmetic series is the sum of an arithmetic sequence. A geometric series is the sum of a geometric sequence. There are other types of series, but you're unlikely to work with them until you're in calculus. For now, you'll probably just work with these two.
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, 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 n = 1, is:
The sum is, in effect, n times the "average" of the first and last terms. This sum of the first n terms is called "the n-th partial sum". (By the way: The above summation formula can be proved using induction.)
The 35th partial sum of this sequence is the sum of the first thirty-five terms. The first few terms of the sequence are:
a1 = (1/2)(1)
+ 1 = 3/2
The terms have a common difference d = 1/2, so this is indeed an arithmetic sequence. The last term in the partial sum will be a35 = a1 + (35 – 1)(d) = 3/2 + (34)(1/2) = 37/2. Then, plugging into the formula, the 35th partial sum is:
(n/2)(a1 + an) = (35/2)(3/2 + 37/2) = (35/2)(40/2) = 350
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. (Plug in values for n if you're not sure about this.) 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?
The quickest way to find the value of
this sum is to find the 14th
partial sums, and then subtract the 14th
from the 47th.
By doing this subtraction, I'll be left with the value of the sum of
terms. The first term is a1
= 2(1) – 5 = –3. The other necessary
terms are a14
= 2(14) – 5 = 23 and a47
= 2(47) – 5 = 89.
Subtracting, I get:
Then the solution is: Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved
By the way, another notation for the summation of the first fourteen terms is "S14", so the subtraction could also be expressed as "S47 – S14".
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:
I know that the first term is 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. Then the summation formula for arithmetical series gives me:
(n/2)(2.25 + [0.25n
+ 2]) = 21
Solving the quadratic, I get that n = –24 (which won't work in this context) or n = 7.
n = 7
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 before you get the right total. So make sure you can do the computations from the formula.
Checking the terms, I can see that this is indeed an arithmetic series: 5 – 1 = 4, 9 – 5 = 4, 53 – 49 = 4. I've got the first and last terms, but how many terms are there in total?
I have the n-th term formula, "an = a1 + (n – 1)d", and I have a1 = 1 and d = 4. Plugging these into the formula, I can figure out how many terms there are:
an = a1
+ (n – 1)d
So there are 14 terms in this series. Now I have all the information I need:
1 + 5 + 9 + ... + 49 + 53 = (14/2)(1 + 53) = (7)(54) = 378