Lesson Arithmetic and Geometric Sequences and Series

A sequence is a set of numbers determined as either arithmetic, geometric, or neither.
Examples:
1.) 1,2,3,4,5,6,7 are all seperated by + 1 ~> Arithmetic
2.) 1,3,9,27,81 are all seperated by * 3 ~> Geometric
3.) 1,4,9,16,25 are neither because they are not seperated by +, -, /, or *
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* Arithmetic Sequence
Equation ~> Tn = T0 + (n - 1)d
Tn = term after "n" units
T0 = first term
n = the amount of numbers after the initial .... for example in (1.), 2 would be the second
d = the common difference .... for example in (1.), all seperated by + 1 .... d = 1
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Now, lets apply this:
Sequence = 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51
What is the 10th number?
tn = t0 + (n - 1)d
t0 = 3
n = 10
d = 4
tn = 3 + (9)4 = 3 + 36 = 39
What is the 13th number?
tn = 3 + (12)4 = 3 + 48 = 51
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* Geometric Sequence
Equation ~> Tn = t0 * r^(n - 1)
t0 = first term
r = common ratio .... for example in (2.), r = 3 .... way to find out ~> divide second number by first or divide third number by second or divide fourth number by third
n = the nth number in the sequence
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Now, lets apply this:
Sequence = 2, 4, 8, 16, 32, 64, 128, 256
What is the 6th number?
t0 = 2
r = 2
n = 6
Tn = 2*2^(5) = 2^6 = 64
What is the 8th number?
Tn = 2*2^(7) = 2^8 = 256
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* Arithmetic Series
Equation ~> Sum = n(t0 + tn)/2
n = number of digits to sum ~> if you are trying to find the sum of 3 numbers ~> n = 3
t0 = first number
tn = last number
Different Version ~>
Sum = n(t0 + t0 + (n - 1)d)/2
Sum = n(2*t0 + (n - 1)d)/2
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Now, we should try to apply:
Sequence = 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 ....
What is the sum of the first 4 digits?
t0 = 0
n = 4
d = 6
Sum = n(2*t0 + (n - 1)d)/2
Sum = 4*(3)*6/2 = 2*3*6 = 36
What is the sum of the first 8 digits?
Sum = n(2*t0 + (n - 1)d)/2
Sum = 8*(7)*6/2 = 4*7*6 = 168
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* Geometric Series
The equation can be quite "breath-taking", but it is my favorite ... :)
Sum = t0 + t0 * r + t0 * r^2 + t0 * r^3 + t0 * r^4 .... + t0 * r^(n - 1)
... reasoning ~> t0 (first term) + t0 * r (second term) ....
Sum * r = t0 * r + t0 * r^2 + t0 * r^3 + t0 * r^4 .... + t0 * r^(n - 1) + t0 * r^(n)
Sum * r - Sum = t0 * r^n - t0
Sum(r - 1) = t0(r^n - 1)
Sum = t0*(r^n - 1)/(r - 1) = t0*(1 - r^n)/(1 - r)
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Now, we should try to apply:
Sequence = 2, 6, 18, 54, 162, 486 ....
What is the sum of the first 4 digits?
t0 = 2
r = ?
6/2 = 3
18/6 = 3
r = 3
n = 4
Sum = t0*(1 - r^n)/(1 - r)
Sum = 2*(1 - 3^4)/(1 - 3)
Sum = 2*(1 - 81)/(-2) = -(1 - 81) = 80
What is the sum of the first 6 digits?
Sum = t0*(1 - r^n)/(1 - r)
Sum = 2*(1 - 3^6)/(1 - 3)
Sum = 2*(1 - 729)/(-2) = -(1 - 729) = 728