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Arithmetic Series (page 4 of 5)

Sections: Terminology and notation, Basic examples, Arithmetic and geometric sequences, Arithmetic series, Finite and infinite geometric series

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.)

• Find the 35th partial sum of a_n = (1/2)n + 1

The 35th partial sum is the sum of the first thirty-five terms. The first few terms are:

a₁ = (1/2)(1) + 1 = 3/2
a₂ = (1/2)(2) + 1 = 2
a₃ = (1/2)(3) + 1 = 5/2

The terms have a common difference d = 1/2, so this is 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 + a_n) = (35/2)(3/2 + 37/2) = (35/2)(40/2) = 350

• Find the value of the following summation:

The quickest way to find this sum is to find the 14th and 47th partial sums, and then subtract the 14th from the 47th, leaving the value of the 15th through 47th 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 2006-2008 All Rights Reserved

• Find the value of n for which the following equation is true:

I know that the first term is a₁ = 0.25(1) + 2 = 2.25. Then the summation formula gives me:

(n/2)(2.25 + [0.25n + 2]) = 21
n(2.25 + 0.25n + 2) = 42
n(0.25n + 4.25) = 42
0.25n² + 4.25n – 42 = 0
n² + 17n – 168 = 0
(n + 24)(n – 7) = 0

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.

• Find the sum of 1 + 5 + 9 + ... + 49 + 53.

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, "a_n = 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:

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:

1 + 5 + 9 + ... + 49 + 53 = (14/2)(1 + 53) = (7)(54) = 378

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