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Arithmetic Series
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SECTION 8-3ARITHMETIC SERIES
WARM-UP
1. FIND THE SUM OF THE FIRST 100 TERMS OF THE ARITHMETIC SEQUENCE 3, 7, 11, ...
2. FIND THE SUM OF THE FIRST 101 TERMS OF THE SAME SEQUENCE.
WARM-UP
1. FIND THE SUM OF THE FIRST 100 TERMS OF THE ARITHMETIC SEQUENCE 3, 7, 11, ...
2. FIND THE SUM OF THE FIRST 101 TERMS OF THE SAME SEQUENCE.
20100
WARM-UP
1. FIND THE SUM OF THE FIRST 100 TERMS OF THE ARITHMETIC SEQUENCE 3, 7, 11, ...
2. FIND THE SUM OF THE FIRST 101 TERMS OF THE SAME SEQUENCE.
20100
20503
SERIES:
SERIES: THE SUM OF A NUMBER OF TERMS OF A SEQUENCE
SERIES: THE SUM OF A NUMBER OF TERMS OF A SEQUENCE
INFINITE SERIES:
SERIES: THE SUM OF A NUMBER OF TERMS OF A SEQUENCE
INFINITE SERIES: A SERIES WITH AN INFINITE NUMBER OF TERMS
SERIES: THE SUM OF A NUMBER OF TERMS OF A SEQUENCE
INFINITE SERIES: A SERIES WITH AN INFINITE NUMBER OF TERMS
FINITE SERIES:
SERIES: THE SUM OF A NUMBER OF TERMS OF A SEQUENCE
INFINITE SERIES: A SERIES WITH AN INFINITE NUMBER OF TERMS
FINITE SERIES: A SERIES WITH A FINITE NUMBER OF TERMS
SERIES: THE SUM OF A NUMBER OF TERMS OF A SEQUENCE
INFINITE SERIES: A SERIES WITH AN INFINITE NUMBER OF TERMS
FINITE SERIES: A SERIES WITH A FINITE NUMBER OF TERMS
ARITHMETIC SERIES:
SERIES: THE SUM OF A NUMBER OF TERMS OF A SEQUENCE
INFINITE SERIES: A SERIES WITH AN INFINITE NUMBER OF TERMS
FINITE SERIES: A SERIES WITH A FINITE NUMBER OF TERMS
ARITHMETIC SERIES: A SERIES WHOSE TERMS ARE TAKEN FROM AN ARITHMETIC SEQUENCE
EXAMPLE 1
EVALUATE.
a. n 2 − 1( )
n=1
8
∑
b. n 2( ) − 1
n=1
8
∑
EXAMPLE 1
EVALUATE.
a. n 2 − 1( )
n=1
8
∑
b. n 2( ) − 1
n=1
8
∑
= 0 + 3 + 8 + 15 + 24 + 35 + 48 + 63
EXAMPLE 1
EVALUATE.
a. n 2 − 1( )
n=1
8
∑
b. n 2( ) − 1
n=1
8
∑
= 0 + 3 + 8 + 15 + 24 + 35 + 48 + 63 = 196
EXAMPLE 1
EVALUATE.
a. n 2 − 1( )
n=1
8
∑
b. n 2( ) − 1
n=1
8
∑
= 0 + 3 + 8 + 15 + 24 + 35 + 48 + 63 = 196
= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 - 1
EXAMPLE 1
EVALUATE.
a. n 2 − 1( )
n=1
8
∑
b. n 2( ) − 1
n=1
8
∑
= 0 + 3 + 8 + 15 + 24 + 35 + 48 + 63 = 196
= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 - 1 = 203
THE GREAT GAUSSHELPING US TO SUM SERIES’
http://www.rare-earth-magnets.com/images2/johann_carl_friedrich_gauss.jpg
THEOREM
THE SUM OF AN ARITHMETIC SERIES WITH FIRST TERM a1 AND CONSTANT DIFFERENCE d IS GIVEN BY:
S
n= n
2a
1+ a
n( )OR
S
n= n
22a
1+ n − 1( )d⎡⎣ ⎤⎦
EXAMPLE 2
A PACKER HAD TO FILL 100 BOXES IDENTICALLY WITH MACHINE TOOLS. THE SHIPPER FILLED THE
FIRST BOX IN 13 MINUTES, BUT GOT FASTER BY THE SAME AMOUNT EACH TIME AS TIME WENT ON. IF HE
FILLED THE LAST BOX IN 8 MINUTES, WHAT WAS THE TOTAL TIME THAT IT TOOK TO FILL THE 100
BOXES?
EXAMPLE 2
A PACKER HAD TO FILL 100 BOXES IDENTICALLY WITH MACHINE TOOLS. THE SHIPPER FILLED THE
FIRST BOX IN 13 MINUTES, BUT GOT FASTER BY THE SAME AMOUNT EACH TIME AS TIME WENT ON. IF HE
FILLED THE LAST BOX IN 8 MINUTES, WHAT WAS THE TOTAL TIME THAT IT TOOK TO FILL THE 100
BOXES?
a1= 13
EXAMPLE 2
A PACKER HAD TO FILL 100 BOXES IDENTICALLY WITH MACHINE TOOLS. THE SHIPPER FILLED THE
FIRST BOX IN 13 MINUTES, BUT GOT FASTER BY THE SAME AMOUNT EACH TIME AS TIME WENT ON. IF HE
FILLED THE LAST BOX IN 8 MINUTES, WHAT WAS THE TOTAL TIME THAT IT TOOK TO FILL THE 100
BOXES?
a1= 13
a100= 8
EXAMPLE 2
A PACKER HAD TO FILL 100 BOXES IDENTICALLY WITH MACHINE TOOLS. THE SHIPPER FILLED THE
FIRST BOX IN 13 MINUTES, BUT GOT FASTER BY THE SAME AMOUNT EACH TIME AS TIME WENT ON. IF HE
FILLED THE LAST BOX IN 8 MINUTES, WHAT WAS THE TOTAL TIME THAT IT TOOK TO FILL THE 100
BOXES?
a1= 13
a100= 8
S100
= 1002
13 + 8( )
EXAMPLE 2
A PACKER HAD TO FILL 100 BOXES IDENTICALLY WITH MACHINE TOOLS. THE SHIPPER FILLED THE
FIRST BOX IN 13 MINUTES, BUT GOT FASTER BY THE SAME AMOUNT EACH TIME AS TIME WENT ON. IF HE
FILLED THE LAST BOX IN 8 MINUTES, WHAT WAS THE TOTAL TIME THAT IT TOOK TO FILL THE 100
BOXES?
a1= 13
a100= 8
S100
= 1002
13 + 8( ) = 1050
EXAMPLE 2
A PACKER HAD TO FILL 100 BOXES IDENTICALLY WITH MACHINE TOOLS. THE SHIPPER FILLED THE
FIRST BOX IN 13 MINUTES, BUT GOT FASTER BY THE SAME AMOUNT EACH TIME AS TIME WENT ON. IF HE
FILLED THE LAST BOX IN 8 MINUTES, WHAT WAS THE TOTAL TIME THAT IT TOOK TO FILL THE 100
BOXES?
a1= 13
a100= 8
S100
= 1002
13 + 8( ) = 1050 MINUTES
EXAMPLE 3
A NEW BUSINESS DECIDES TO RANK ITS 9 EMPLOYEES BY HOW WELL THEY WORK AND PAY
THEM AMOUNTS THAT ARE IN AN ARITHMETIC SEQUENCE WITH A CONSTANT DIFFERENCE OF $500
A YEAR. IF THE TOTAL AMOUNT PAID THE EMPLOYEES IS TO BE $250,000, WHAT WILL THE
EMPLOYEES MAKE PER YEAR?
EXAMPLE 3
A NEW BUSINESS DECIDES TO RANK ITS 9 EMPLOYEES BY HOW WELL THEY WORK AND PAY
THEM AMOUNTS THAT ARE IN AN ARITHMETIC SEQUENCE WITH A CONSTANT DIFFERENCE OF $500
A YEAR. IF THE TOTAL AMOUNT PAID THE EMPLOYEES IS TO BE $250,000, WHAT WILL THE
EMPLOYEES MAKE PER YEAR?
S
9= 250,000 = 9
22a
1+ 9 − 1( )500⎡⎣ ⎤⎦
EXAMPLE 3
A NEW BUSINESS DECIDES TO RANK ITS 9 EMPLOYEES BY HOW WELL THEY WORK AND PAY
THEM AMOUNTS THAT ARE IN AN ARITHMETIC SEQUENCE WITH A CONSTANT DIFFERENCE OF $500
A YEAR. IF THE TOTAL AMOUNT PAID THE EMPLOYEES IS TO BE $250,000, WHAT WILL THE
EMPLOYEES MAKE PER YEAR?
S
9= 250,000 = 9
22a
1+ 9 − 1( )500⎡⎣ ⎤⎦
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
500,000
9= 2a
1+ 4000
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
464,000
9= 2a
1
500,000
9= 2a
1+ 4000
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
464,000
9= 2a
1
500,000
9= 2a
1+ 4000
232000
9= a
1
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
464,000
9= 2a
1
500,000
9= 2a
1+ 4000
232000
9= a
1 ≈ $25778
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
464,000
9= 2a
1
500,000
9= 2a
1+ 4000
232000
9= a
1 ≈ $25778
a
9≈ 25778 + 9 − 1( )500
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
464,000
9= 2a
1
500,000
9= 2a
1+ 4000
232000
9= a
1 ≈ $25778
a
9≈ 25778 + 9 − 1( )500 ≈ 25778 + 4000
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
464,000
9= 2a
1
500,000
9= 2a
1+ 4000
232000
9= a
1 ≈ $25778
a
9≈ 25778 + 9 − 1( )500 ≈ 25778 + 4000 ≈ $29778
EXAMPLE 3
250,000 = 9
22a
1+ 4000⎡⎣ ⎤⎦
464,000
9= 2a
1
500,000
9= 2a
1+ 4000
232000
9= a
1 ≈ $25778
a
9≈ 25778 + 9 − 1( )500 ≈ 25778 + 4000 ≈ $29778
THE EMPLOYEES WILL EARN BETWEEN $25,778 AND $29,778 PER YEAR
HOMEWORK
HOMEWORK
P. 507 #1-22
