13.3 Arithmetic and Geometric Series and Their Sums Finite Series

An arithmetic series is the sum of an arithmetic sequence 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 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.

Arithmetic Sequence The formula for the first n terms of an arithmetic sequence, starting with n = 1, is:

a1 = (1/2)(1) + 1 = 3/2 a2 = (1/2)(2) + 1 = 2 a3 = (1/2)(3) + 1 = 5/2 Find the 35th partial sum of an = (1/2)n + 1 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 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 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 or a35 = (1/2)(35) + 1 = 35/2 + 1 = 37/2 Then, plugging into the formula, the 35th partial sum is:

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, "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 53 = 1 + (n – 1)(4) 53 = 1 + 4n – 4 53 = 4n – 3 56 = 4n 14 = n

Find the value of the following summation:            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, leaving the value of the sum of the 15th through 47th terms.

Find the sum of all positive 3 digit integers that are divisible by 8. The first 3 digit integer divisible by 8 is 104 and the last 3 digit integer divisible by 8 is 992. We need to find n, the number of 3 digit integers that are divisible by 8.

Now find S112

Geometric Sequence For a geometric sequence with first term a1 and common ratio r, the sum of the first n terms is given by:

Evaluate the following:           The first few terms are –6, 12, –24, so this is a geometric series with common ratio r = –2. The first term of the sequence is a = –6.

Plugging into the summation formula, I get

Evaluate S10 for 250, 100, 40, 16,.... I need to find the sum of the first ten terms. The first term is a = 250. Dividing pairs of terms, I get 100 ÷ 250 = 2/5, 40 ÷ 100 = 2/5, etc, so the terms being added form a geometric sequence with common ratio r = 2/5 416.62

Assignment Page 489 #1-8,27-32