Andrew M. answered • 06/12/15

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Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

For an arithmetic sequence each subsequent term is found by adding a constant to the previous term

This constant is called the difference or 'd'

The formula a

_{n}= a_{1}+ (n-1)d indicates that we added the difference d to the original term (n-1) timesExample: Find the 10th term of sequence 4, 7, 10, 13, 16, ...

We see that d = 3 because each new term is found by

adding 3 to the previous term.

To find the 10th term in this sequence we would have a

_{1}= 4, d = 3, n-1 = 9a

_{10}= 4 + (10-1)(3) = 4 + (9)(3) = 31********************************

S

_{n}=n(a_{1}+a_{n})/2This formula is for finding the sum of the first n terms of an arithmetic sequence.

This means we add a

_{1}+ a_{2}+ a_{3}+ ... + a_{n}For example:

Find the sum of the first 12 terms of the arithmetic sequence a

_{n}= 2n + 4First we determine this is an arithmetic sequence and find a

_{1}and d so we canuse the formula a

_{n}= a_{1}+ (n-1)d to find a_{12}a

_{1}= 2(1) + 4 = 6a

_{2}= 2(2) + 4 = 8a

_{3}= 2(3) + 4 = 10We see that this is an arithmetic sequence where d = 2

We can find a

_{12}from the formula a_{n}= a_{1}+ (n-1)da

_{12}= 6 + (12-1)(2) = 6 + 22 = 28We now have a

_{1}= 6, a_{12}= 28, n = 12 so we plug into S_{n}=n(a_{1}+a_{n})/2S

_{12}= 12(6+28)/2 = 12(34)/2 = 6(34) = 204This means that the first twelve terms of this arithmetic sequence add to 204

*************************************

**a**

_{n}=a_{1}(r)^{n-1}A geometric sequence is where each subsequent term is found by multiplying the

previous term by a set number called the common ratio..., a

_{n+1}= r(a_{n})For example 2, 6, 18, 54, 162...

In this geometric sequence r=3 because each subsequent term is found by

multiplying the previous term by 3 ..

Example: Find the 11th term of the geometric sequence 2, 6, 18, 54, ...

As already stated r=3 a

_{1}= 2 n = 11a

_{11}= 2(3)^{11-1}= 2(3)^{10}= 118098*********************************

**Sn=a**

_{1}(1-r^{n})/(1-r)This geometric series is basically telling you how to find the sum of the

first n terms of a geometric sequence.

Example: Let's look at the previously used geometric sequence 2, 6, 18, 54, 162...

Find the sum of the first 8 terms of the geometric sequence 2, 6, 18, 54, 162...

We have: a

_{1}= 2 r = 3 n = 8 so plug into the formulaS

_{8}= 2(1-(3^{8}))/(1-3) = 2(1-3^{8})/(-2) = -1(1-6561) = -1(-6560) = 6560Thus the sum of the first 8 terms of the geometric sequence is 6560

**********************************

S

_{∞}= a_{1}/(1-r)This is the formula to find out what the sum of a geometric sequence approaches as

the number of entries in the sequence approaches infinity.

This sum will exist if the common ratio r is in the range -1 < r < 1

Or absolute value of r is less than 1. This basically means that the

common ratio is a fraction less than 1 and greater than -1

Example: Find S

_{∞ }[4(1/2)^{n-1}]This is set up in the format of a geometric sequence a

_{n}= a_{1}(r)^{n-1}We have a

_{1}= 4 r = 1/2Since r is a fraction between -1 and 1 the sum exists so plug into the formula

S

_{∞}= 4/(1-(1/2)) = 4/(1/2) = 4(2) = 8This means that the sum of the terms in this geometric sequence will approach 8

as the number of terms approaches infinity

Andrew M.

You're welcome Rob. Hope this clears it up some.

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06/12/15

Rob D.

06/12/15