SCRIBE: Sequences & Series

Well Michael ended off his scribe with a coded sequence to reveal who the next scribe with so I'll show you how we found out it was me. 14, 34, 39, 43, 47, 56, 61.

The differences between each numbers are as follows

20, 5, 4, 4, 9, 5

Since he told us that the differences represented the order of the alphabets, we get this word.

T E D D I E

So today's scribe is me Teddie!

Okay enough with the fun and game and lets get down to the stuff we did today cuz it was a whole lot.

THE MATH DICTIONARY

SEQUENCES & SERIES

Sequence: An ordered list of numbers that follows a certain rule or pattern.

Series: The sum of numbers in a sequence to a particular term.

Example: Given the sequence
3, 6, 12, 24, 48, . . .

S₁ = 3
S₂ = 3 + 6 = 9
S₃ = 3 + 6 + 12 = 21
S₄ = 3 + 6 + 12 + 24 = S₅ = 3 + 6 + 12 + 24 + 48 = S_n = 3 + 6 + 12 + 24 + 48 + . . . . t_n
S₅ Denotes the sum of the first 5 terms.
S_n denotes the sum of the first n terms.

ARITHMETIC SEQUENCE

I ) Recursive Definition: an ordered list of numbers generated by continually adding a value (the common difference) to a given first term.

II ) Implicit Definition: an ordered list of numbers where each number in the list is generated by a linear equation.

III ) Common difference (d):
a) the number that is repeatedly added to successive terms in an arithmetic sequence.
b) from the implicit definition d is the slope of the linear equation.

To Find a Common Difference

d = t_n -t_(n-1)

d is the common difference
t_n is any term in the sequence other than the first term.
t₁ is the first term in the sequence

Finding the n^the Term in an Arithmetic Sequence

t_n = t₁ + (n - 1)d

t_n is the n^the term in the sequence
t₁ is the first term n is the rank of the term
d is the common difference

Example: find the 51^>st term of the arithmetic sequence 11, 5, -1, -7, . . . .

Solution:

n = 51
t₁ = 11
d = -6

t₅₁ = t₁ + (n - 1)d
= 11 + (51 - 1)(-6)
= 11 + (50)(-6)
= 11 - 300
= -289

ARITHMETIC MEANS

The term(s) that falls between any two non-consecutive terms in an arithmetic sequence

Examples: Find the arithmetic means in each sequence

a) 1, _ , 25
there are 2 gaps for differences to occur
25 - 1 = 24
2d = 24
d = 12

1, 13 , 25

b) 3, _ , _ , _ , 1
there are 4 gaps for differences to occur
1 - 3 = -2
4d = -2
d = -0.5

3, 2.5, 2, 1.5, 1

Arithmetic Series: the sum of numbers in an arithmetic sequence given by:

S_n = (n/2) [ 2t₁ + (n - 1) d ]

d is the common difference
S_n is the sum to the n^the term
t₁ is the first term

Example:
find the sum to the 21^st term of the sequence
5, 9, 13, 17, . . .

Solution:
t₁ = 5
d = 4
n = 21

S_n = (n/2) [ 2t₁ + (n - 1) d ]
S₂₁ = (21/2) [ 2(5) + (21 - 1)(4) ]
S₂₁ = (21/2) [ 10 + 80 ]
S₂₁ = (21/2) [ 90 ]
S₂₁ = 21 [ 45 ]
S₂₁ = 945

GEOMETRIC SEQUENCES:

I ) Recursive: an ordered list of numbers generated by repeatedly multiplying a given first term by a particular value (the common ratio).

II ) Implicit: the ordered list of numbers where each number in the list is generated by an exponential function

And that is all for the notes in our dictionary, pew.

Starting off the afternoon we started off with some problems.

1. Find S_n for each arithmetic sequence

a) t₁ = 8
d = 5
S₂₅ = ?

S₂₅ = (25/2) [ 2(8) + (25 - 1) 5 ]
S₂₅ = (25/2) [ 16 + 120 ]
S₂₅ = (25/2) [ 136 ]
S₂₅ = (25) [68]
S₂₅ = 1700

b) t₅ = 15
t₁₉ = 43
S₁₀₀ = ?
d = ?
t₁ = ?

19 - 5 = 14 common differences required
43 - 15 = 28
14d = 28
d = 2

t₅ = t₁ + (5 - 1) d
15 = t₁ + (4)(2)
15 = t₁ + 8
7 = t₁

S₁₀₀ = (100/2) [ 2(7) + (100 - 1)(2) ]
S₁₀₀ = 50 [ 14 + 198 ]
S₁₀₀ = = 50 [212]
S₁₀₀ = 10600

2. Given S₁₂ = 360, t₁ = -3 find d
S₁₂ = (12/2) [ 2 t₁ + (12 - 1) d ]
360 = 6 [ 2(-3) + 11d ]
360 = -36 + 66d
396 = 66d
d = 6

3. Given S₃₀ = 480 t₁ = 4 find t₃₀

S₃₀ = (30/2) [ 2t_{1 + (30 - 1) d ]} 480 = 15 [ 2(4) + 29d ]
480 = 120 + 435d
360 = 435d
24/29 = d

t₃₀ = t₁ + (30 - 1) d
t₃₀ = 4 + 29(24/29)
t₃₀ = 4 + 24
t₃₀ = 28

4. x = 1, (1/2)x + 4, 1 - 2x form and arithmetic sequence what are the first 3 terms?

[(1/2)x + 4] - (x - 1) = (1 - 2x) - [(1/2)x + 4]
1/2)x + 4 - x - 1 = 1 - 2x - (1/2)x + 4
2x = -8
x = -4
x - 1 = -4 -1 = 5
(1/2)x + 4 = (1/2)(-4) + 4 = 2 1 - 2x = 1 - 2(-4) = 9

Finally I am done yay! And tomorrow's scribe is CALVIN!