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Chapter 11: Test Your Proficiency. Directions: Select a section to work on. Work out each problem on a piece of paper. Click to check your answer. For detailed steps click on the provided link. Move on to the next problem or return to the menu. Sections. 11-1: The Arithmetic Sequence. - PowerPoint PPT Presentation
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Chapter 11: Test Your Proficiency
Directions: •Select a section to work on.•Work out each problem on a piece of paper. •Click to check your answer.•For detailed steps click on the provided link.•Move on to the next problem or return to the menu.
Sections
11-1: The Arithmetic Sequence
11-2: The Arithmetic Series
11-3: The Geometric Sequence
11-4: The Geometric Series
11-5: The Infinite Geometric Series
11-7: The Binomial Theorem
11-8: Mathematical Induction
11-6: Recursion and Iteration
I love sequences and
series!
Applications/Word Problems
11-1: The Arithmetic SequenceGo on to the next problem
Check Answer
1. Find the 10th term of the arithmetic sequence
9,13,17, Use the correct formula.
Return to
Menu
11-1: The Arithmetic Sequence
Click here if you would you like to see a detailed explanation.
10 45a
1. Find the 10th term of the arithmetic sequence
9,13,17, Use the correct formula.
Return to
Menu
Go on to the next problem
11-1: The Arithmetic Sequence
Step 1: Recall the formula to find a term of an arithmetic sequence. 1 1na a n d Step 2: Identify the known values and plug them into the formula.
10
1
1
0
4
1
9 9
10
3 4
?
10
9
1
a
a
a
n
d
Step 3: Simplify and state the answer.
10
10
10
1
9 9 4 9 36
4
109 4
5
a
a
a
1. Find the 10th term of the arithmetic sequence
9,13,17, Use the correct formula.
Return to
Menu
Go on to the next problem
11-1: The Arithmetic SequenceGo on to the next problem
Check Answer
2. Find the three arithmetic means between
18 and 46. Use the correct formula.
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11-1: The Arithmetic Sequence
The arithmetic means are 25,32,39.
2. Find the three arithmetic means between
18 and 46. Use the correct formula.
Click here if you would you like to see a detailed explanation.
Return to
Menu
Go on to the next problem
11-1: The Arithmetic Sequence
Step 1: Set up the problem and recall the formula to find a term of an arithmetic sequence.
118, ____, ____, ____,46 formula : 1 na a n dStep 2: Identify the known values and plug them into the formula.
1 18 1
46
46 8 5 1
?
5
na
a d
d
n
Step 3: Solve for d. Then find the missing arithmetic means by adding d to each preceding term.
46 18 5 1
46 18 4
28 4
7
d
d
d
d
2
3
4
18 7 25
25 7 32
32 7 39
a
a
a
The arithmetic means are 25,32,39.
2. Find the three arithmetic means between
18 and 46. Use the correct formula.
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Menu
Go on to the next problem
11-1: The Arithmetic SequenceGo on to the next problem
Check Answer
3. 230 is the ____ term of the arithmetic sequence
10, 2, 14, ? Use the correct formula.
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Menu
11-1: The Arithmetic Sequence
It is the 21st term.
3. 230 is the ____ term of the arithmetic sequence
10, 2, 14, ? Use the correct formula.
Click here if you would you like to see a detailed explanation.
Return to
Menu
Go on to the next problem
11-1: The Arithmetic Sequence
Step 1: Recall the formula to find a term of an arithmetic sequence. 1 1na a n d
Step 2: Identify the known values and plug them into the formula.
1
12
2 10 12
230 10
10
30 12
?
n n
n
d
a
a
Step 3: Solve for the missing value and state the answer.
230 10 1 12
240 12 1
20 1
21
n
n
n
n
3. 230 is the ____ term of the arithmetic sequence
10, 2, 14, ? Use the correct formula.
It is the 21st term.
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Menu
Go on to the next problem
11-1: The Arithmetic SequenceGo on to the next problem
Check Answer
4. 270 is the ____ term of the arithmetic sequence
10,4,18, ? Use the correct formula.
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Menu
11-1: The Arithmetic Sequence
It is the 21st term.
4. 270 is the ____ term of the arithmetic sequence
10,4,18, ? Use the correct formula.
Click here if you would you like to see a detailed explanation.
Return to
Menu
Go on to the next problem
11-1: The Arithmetic Sequence
Step 1: Recall the formula to find a term of an arithmetic sequence. 1 1na a n d
Step 2: Identify the known values and plug them into the formula.
1
1410
4 10
1
270 270
4
1
1
0
?n n
n
a
a
d
Step 3: Solve for the missing value and state the answer.
270 10 1 14
280 14 1
20 1
21
n
n
n
n
It is the 21st term.
4. 270 is the ____ term of the arithmetic sequence
10,4,18, ? Use the correct formula.
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Go on to the next problem
11-1: The Arithmetic SequenceGo on to the next problem
Check Answer
5. Find the 27th term of the arithmetic sequence
16,9,2, Use the correct formula.
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11-1: The Arithmetic Sequence
Click here if you would you like to see a detailed explanation.
27 166a
5. Find the 27th term of the arithmetic sequence
16,9,2, Use the correct formula.
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Menu
Go on to the next problem
11-1: The Arithmetic Sequence
Step 1: Recall the formula to find a term of an arithmetic sequence. 1 1na a n d Step 2: Identify the known values and plug them into the formula.
1 27
27
16 1
27
27 1
?
7
9 1
6
6 7
a
a
a
d
n
Step 3: Simplify and state the answer.
27
27
27
1
16 26
16
7 16 182
1
727
66
a
a
a
5. Find the 27th term of the arithmetic sequence
16,9,2, Use the correct formula.
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Go on to the next problem
11-2: The Arithmetic SeriesGo on to the next problem
1. Find the sum of the first 12 terms of the arithmetic series
21 29 37 .
Check Answer
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11-2: The Arithmetic Series
1. Find the sum of the first 12 terms of the arithmetic series
21 29 37 .
The sum is -780.
Click here if you would you like to see a detailed explanation.
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Go on to the next problem
11-2: The Arithmetic Series
1. Find the sum of the first 12 terms of the arithmetic series
21 29 37 .
Step 1: Recall the formula to find the sum of an arithmetic series. 12n nn
S a a
Step 2: Identify the known values and unknown values.
1 1221 12 ? 8 ?na n a d S
Step 3: Find the value of an.
1 1 21 12 1
109
8n n
n
a a n d a
a
Step 4: Find Sn.
12 12
12
1221 109 6 130
2780
S S
S
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11-2: The Arithmetic SeriesGo on to the next problem
n
1
2. Find S for the arithmetic series in which
8, 161,and 52.na a n
Check Answer
Return to
Menu
11-2: The Arithmetic Series
The sum is -4394.
Click here if you would you like to see a detailed explanation.
n
1
2. Find S for the arithmetic series in which
8, 161,and 52.na a n
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Go on to the next problem
11-2: The Arithmetic Series
Step 1: Recall the formula to find the sum of an arithmetic series. 12n nn
S a a
Step 2: Identify the known values and unknown values.
1 528 52 161 ?na n a S
Step 3: Plug in the values to the formula and find Sn.
52 52
52
528 161 26 169
24394
S S
S
n
1
2. Find S for the arithmetic series in which
8, 161,and 52.na a n
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Go on to the next problem
11-2: The Arithmetic SeriesGo on to the next problem
3. Find the sum of the arithmetic series -16 + -11 + -6 + + 54.
Check Answer
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11-2: The Arithmetic Series
The sum is 285.
Click here if you would you like to see a detailed explanation.
3. Find the sum of the arithmetic series -16 + -11 + -6 + + 54.
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11-2: The Arithmetic Series
3. Find the sum of the arithmetic series -16 + -11 + -6 + + 54.
Step 1: Recall the formula to find the sum of an arithmetic series. 12n nn
S a a
Step 2: Identify the known values and unknown values.
1 16 ? 54 11 ( 16) 5 ?n na n a d S
Step 3: Find the value of n.
1Use 1 54 16 1 5
70 5 1
14 1
15
na a n d n
n
n
n
Step 4: Find Sn.
15 15
15
1516 54 7.5 38
2285
S S
SReturn
to Menu
Go on to the next problem
11-2: The Arithmetic SeriesGo on to the next problem
18
3
4. Find the sum of the arithmetic series 3 4 . n
n
Check Answer
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Menu
11-2: The Arithmetic Series
The sum is 568.
Click here if you would you like to see a detailed explanation.
18
3
4. Find the sum of the arithmetic series 3 4 . n
n
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Go on to the next problem
11-2: The Arithmetic Series
Step 1: Recall the formula to find the sum of an arithmetic series. 12n nn
S a a
Step 2: Find a1.
1
1
For the value of (the first n value).
So, a
n is 3
33 4 13
a
18
3
4. Find the sum of the arithmetic series 3 4 . n
n
Step 3: Find an.
Step 4: Find the value of n.
n
For the value of (the last n value).
So, a 3
n is 18
18 4 58
na
For the value of
18 3 1 16
(always subtract the last and first n values and ad )
n:
d 1
n
Click to see the rest of the explanation.
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11-2: The Arithmetic Series
Step 5: Plug in the values and find Sn.
16 16
16
1613 58 8 71
2568
S S
S
18
3
4. Find the sum of the arithmetic series 3 4 . n
n
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11-3: The Geometric SequenceGo on to the next problem
1
1. Find the ninth term of the geometric sequence for which
2a 20 and . (Use the appropriate formula.)
3r
Check Answer
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11-3: The Geometric Sequence
Click here if you would you like to see a detailed explanation.
1
1. Find the ninth term of the geometric sequence for which
2a 20 and . (Use the appropriate formula.)
3r
95120
6561a
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11-3: The Geometric Sequence
Step 1: Recall the formula to find a term of a geometric sequence. 11
nna a r
Step 2: Plug in the given values.
Step 3: Simplify.
Step 4: State the answer.
1
1. Find the ninth term of the geometric sequence for which
2a 20 and . (Use the appropriate formula.)
3r
9 1
92
Note: n = 9 (for ninth term) 203
a
8
9 8
20 2 20 256 5120
1 1 6561 65613a
95120
6561a
Note: Do not use rounded decimals unless the directions tell you to.
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11-3: The Geometric SequenceGo on to the next problem
1
2. Find the eleventh term of the geometric sequence for which
a 13 and 3. (Use the appropriate formula.)r
Check Answer
Return to
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11-3: The Geometric Sequence
Click here if you would you like to see a detailed explanation.
11 767,637a
1
2. Find the eleventh term of the geometric sequence for which
a 13 and 3. (Use the appropriate formula.)r
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Menu
Go on to the next problem
11-3: The Geometric Sequence
Step 1: Recall the formula to find a term of a geometric sequence. 11
nna a r
Step 2: Plug in the given values.
Step 3: Simplify.
Step 4: State the answer.
11 111Note: n = 11 (for eleventh term) 13 3a
1011 13 3 767,637a
Note: always enclose negative numbers in parentheses when raising to a power
1
2. Find the eleventh term of the geometric sequence for which
a 13 and 3. (Use the appropriate formula.)r
11 767,637a
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11-3: The Geometric SequenceGo on to the next problem
Check Answer
27
3. Find the next three terms of the geometric sequence 6, 9, , .2
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11-3: The Geometric Sequence
Click here if you would you like to see a detailed explanation.
81 243 729, ,
4 8 16
27
3. Find the next three terms of the geometric sequence 6, 9, , .2
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11-3: The Geometric Sequence
Strategy: Use the definition of a geometric sequence: Each term after the first is found by multiplying the previous term by the common ration.
Step 1: Find the value of r (the common ratio).
Step 2: Find the next three terms in the sequence. Step 3: State the answer.
2
1
9 3Recall:
6 2
ar r
a
4 3
5 4
6 5
27 3 81
2 2 4
81 3 243
4 2 8
243 3 729
8 2 16
a a r
a a r
a a r
27
3. Find the next three terms of the geometric sequence 6, 9, , .2
81 243 729, ,
4 8 16
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11-4: The Geometric SeriesGo on to the next problem
Check Answer
7 7 14
1. Find the sum of the geometric series to 8 terms.6 3 3
Use the appropriate formula.
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Click here if you would you like to see a detailed explanation.
81 595
297 or 2 2
S
11-4: The Geometric Series
7 7 14
1. Find the sum of the geometric series to 8 terms.6 3 3
Use the appropriate formula.
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Go on to the next problem
Strategy: Use the formula to find the sum of a geometric series with a finite number of terms, n.
Step 1: Find the value of r (the common ratio), the value of a1, and the value of n.
Step 2: Plug in r, a1 and n to the formula. Step 3: State the answer.
21
1
77 6 73 2 8 (the number of terms being added)
7 3 7 66
ar r a n
a
8
8
7 72
6 61 2
S
11-4: The Geometric Series
7 7 14
1. Find the sum of the geometric series to 8 terms.6 3 3
Use the appropriate formula.
81 595
297 or 2 2
S
1 1Recall:
1
n
na a r
Sr
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11-4: The Geometric SeriesGo on to the next problem
Check Answer
16
n=1
12. Find the sum of the geometric series 5 . Use the appropriate formula.
2
n
Return to
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Click here if you would you like to see a detailed explanation.
627 315
9 or 32 32
S
11-4: The Geometric Series
16
n=1
12. Find the sum of the geometric series 5 . Use the appropriate formula.
2
n
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Go on to the next problem
Strategy: Use the formula to find the sum of a geometric series with a finite number of terms, n.
Step 1: Identify the value of r (the common ratio), the value of a1, and the value of n from the sigma notation.
Step 2: Plug in r, a1 and n to the formula.
Step 3: State the answer.
6
6
15 5
21
12
S
11-4: The Geometric Series
1 1Recall:
1
n
na a r
Sr
16
n=1
12. Find the sum of the geometric series 5 . Use the appropriate formula.
2
n
627 315
9 or 32 32
S
1
6 11
52
n
1a r
6 1 1
6
n
n
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11-4: The Geometric SeriesGo on to the next problem
Check Answer
13. Find the sum of the geometric series for which 19, 4, and 10.
Use the appropriate formula.
a r n
Return to
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Click here if you would you like to see a detailed explanation.
10 3,984,585S
11-4: The Geometric Series
13. Find the sum of the geometric series for which 19, 4, and 10.
Use the appropriate formula.
a r n
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Strategy: Use the formula to find the sum of a geometric series with a finite number of terms, n.
Step 1: Plug in r, a1 and n to the formula. Step 2: State the answer.
10
1019 19 4
1 4S
11-4: The Geometric Series
1 1Recall:
1
n
na a r
Sr
10 3,984,585S
Note: always enclose negative numbers in parentheses when raising to a power
13. Find the sum of the geometric series for which 19, 4, and 10.
Use the appropriate formula.
a r n
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Go on to the next problem
11-5: The Infinite Geometric SeriesGo on to the next problem
Check Answer
1. Find the sum of the infinite geometric series, if it exists.
Use the appropriate formula.
8 20 50
Return to
Menu
Click here if you would you like to see a detailed explanation.
51
2r
11-5: The Infinite Geometric Series
1. Find the sum of the infinite geometric series, if it exists.
Use the appropriate formula.
8 20 50
The sum does not exist.
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Recall that an infinite geometric series only has a sum under certain conditions. The value of the common ratio, r, must be within the interval -1 < r < 1. Otherwise, the sum does not exist.
Step 1: Find the value of r. Step 2: Compare the value of r to the interval -1 < r < 1.
2
1
20 5
8 2
ar
a
51
2r
11-5: The Infinite Geometric Series
1. Find the sum of the infinite geometric series, if it exists.
Use the appropriate formula.
8 20 50
The value of r is outside of the interval.
Step 3: Do not use the sum formula. Make the concluding statement.
51
2r The sum does not exist.
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11-5: The Infinite Geometric SeriesGo on to the next problem
Check Answer
2. Find the sum of the infinite geometric series, if it exists.
Use the appropriate formula.
2512 10
3
Return to
Menu
Click here if you would you like to see a detailed explanation.
72S
11-5: The Infinite Geometric Series
2. Find the sum of the infinite geometric series, if it exists.
Use the appropriate formula.
2512 10
3
Return to
Menu
Go on to the next problem
Recall that an infinite geometric series only has a sum under certain conditions. The value of the common ratio, r, must be within the interval -1 < r < 1. Otherwise, the sum does not exist.
Step 1: Find the value of r. Step 2: Compare the value of r to the interval -1 < r < 1.
2
1
10 5
12 6
ar
a
51 1
6
11-5: The Infinite Geometric Series
The value of r is inside of the interval.
Step 3: Use the sum formula for an infinite geometric series.
2. Find the sum of the infinite geometric series, if it exists.
Use the appropriate formula.
2512 10
3
1 12 1212 6 72
5 11 16 6
aS
r
Step 4: Make the concluding statement.
72S
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11-5: The Infinite Geometric SeriesGo on to the next problem
Check Answer
1
1
3. Find the sum of the infinite geometric series, if it exists.
2 Use the appropriate formula. 3
3
n
n
Return to
Menu
Click here if you would you like to see a detailed explanation.
9S
11-5: The Infinite Geometric Series
1
1
3. Find the sum of the infinite geometric series, if it exists.
2 Use the appropriate formula. 3
3
n
n
Return to
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Go on to the next problem
Recall that an infinite geometric series only has a sum under certain conditions. The value of the common ratio, r, must be within the interval -1 < r < 1. Otherwise, the sum does not exist.
Step 1: Identify the value of r from the sigma notation.
Step 2: Compare the value of r to the interval -1 < r < 1.
2
1 13
11-5: The Infinite Geometric Series
The value of r is inside of the interval.
Step 3: Use the sum formula for an infinite geometric series. Note: here a1 is 3
12
3
33 3
13
39
11
aS
r
Step 4: Make the concluding statement.
9S
1
1
3. Find the sum of the infinite geometric series, if it exists.
2 Use the appropriate formula. 3
3
n
n
1
1
23
3
n
nr
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11-6: Recursion and IterationGo on to the next problem
Check Answer
1
1
1. Find the first five terms of the sequence where 7
and 3 8.n n
a
a a
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Menu
Click here if you would you like to see a detailed explanation.
7,13,31,85,247
11-6: Recursion and Iteration
1
1
1. Find the first five terms of the sequence where 7
and 3 8.n n
a
a a
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Step 1: Always begin with n = 1.
Step 2: Now plug in the value of .
Step 3: Repeat this process with n = 2, then n = 3, and then n = 4.
11-6: Recursion and Iteration
1
1
1. Find the first five terms of the sequence where 7
and 3 8.n n
a
a a
7,13,31,85,247
1 1
1
1 1
2
If , then 3 8 becomes 3 8
which yields
n = 1
3 8n na a a a
a a
1a
12 23 8 become 1s 3 7a 8 3a a
1
1
1
2
3
2 3 3
4 4
54 54
2
3 3
4
If , then 3 8 3 8 3 8
If , then 3 8 3
n = 2
n = 8 3 8
If , then 3 8
13
31
8
3
n =
31
85
3 2484 5 73 8
a a a a
a a a a
a a a
a
a
a
a
Step 4: State the first five terms in sequence.
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11-6: Recursion and IterationGo on to the next problem
Check Answer
1 2
2 1
2. Find the first five terms of the sequence where 9, 15
and 2 .n n n
a a
a a a
Return to
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Click here if you would you like to see a detailed explanation.
9,15, 3, 33, 27
11-6: Recursion and Iteration
1 2
2 1
2. Find the first five terms of the sequence where 9, 15
and 2 .n n n
a a
a a a
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Step 1: Always begin with n = 1.
Step 2: Now plug in the values of . Be careful to place the values correctly.
Step 3: Repeat this process with n = 2 and n = 3.
11-6: Recursion and Iteration
1
2 1 11 2
3
1
2
1If , then 2 becomes 2
which yield
n
s 2
= 1 n n na a
a a
a a a
a
a
12 an d a a
23 1 32 become 3s 15a 2 9a a a
2
3
4 42 2 2
3 3
2 1
2 451
3
3 5
If , then 2 2 a 2
I
1
f ,
n 3
then
3
27
3
5
3
= 2
n 2 3= 2 23 a 3
a
a
aa a a
a
a
a a a a
Step 4: State the first five terms in sequence.
1 2
2 1
2. Find the first five terms of the sequence where 9, 15
and 2 .n n n
a a
a a a
9,15, 3, 33, 27 Return to
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11-6: Recursion and IterationGo on to the next problem
Check Answer
03. Find the first three iterates of 9 2 for an initial value of x 5.f x x
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Click here if you would you like to see a detailed explanation.
43, 385, 3463
11-6: Recursion and Iteration
03. Find the first three iterates of 9 2 for an initial value of x 5.f x x
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Step 1: Find the first iterate, x1 by finding f (x0).
Step 2: Find the second iterate, x2, by finding f (x1).
11-6: Recursion and Iteration
1 0 5 2 39 5 4xfx f
Step 4: State the first three iterates in sequence.
03. Find the first three iterates of 9 2 for an initial value of x 5.f x x
43, 385, 3463
12 943 4 383 52f fxx
Step 3: Find the third iterate, x3, by finding f (x2).
23 385 39 2 385 463xfx f
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11-6: Recursion and IterationGo on to the next problem
Check Answer
204. Find the first three iterates of 3 2 for an initial value of x 4.f x x
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46, 6346, 120,815,146
11-6: Recursion and Iteration
204. Find the first three iterates of 3 2 for an initial value of x 4.f x x
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Step 1: Find the first iterate, x1 by finding f (x0). Remember to follow the order of operations.
Step 2: Find the second iterate, x2, by finding f (x1).
11-6: Recursion and Iteration
201 4 4 63 42xx f f
Step 4: State the first three iterates in sequence.
122
3 2 64 46 466 3xfx f
Step 3: Find the third iterate, x3, by finding f (x2).
232
6346 63 120,3 82 1 , 66 144 5f x fx
204. Find the first three iterates of 3 2 for an initial value of x 4.f x x
46, 6346, 120,815,146
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11-7: The Binomial TheoremGo on to the next problem
Check Answer
51. Expand 5 2 using the Binomial Theorem.x
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11-7: The Binomial Theorem
51. Expand 5 2 using the Binomial Theorem.x
5 5 4 3 25 2 3125 6250 5000 2000 400 32x x x x x x
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Note: These problems are much easier if you learn the pattern from term to term rather than try to memorize the Binomial Theorem as a formula.
Step 1: Write out the expansion.
Note the patterns: 1) the powers on the first term (5x) decrease (5, 4, 3, 2, 1, 0)2) The powers on the second term (2) increase (0,1, 2, 3, 4, 5)3) The factorial increases (0!, 1!, 2!, 3!, 4!, 5!)4) The previous power of (5x) moves to the front as a multiplier in the next term.
5 4 3 25
1 2 3
4 51
0! 1
5 5 4 55 5 2 5 2 5 25 2
5 2
4 3
5 4 3 2 5 4 3
! 2!
2
3!
4
1 2
! 5!
x x x xx
x
Continue to the next slide to see the rest of the explanation.
11-7: The Binomial Theorem
51. Expand 5 2 using the Binomial Theorem.x
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Step 2: Simplify the terms in the expansion. Remember to follow the order of operations.
5 4 3 25
3125 625 2 125 4 25 8
1 1 2
5 5 4 5 4 3
5 4 3 2 5 4 3 2
5 2
5 16 11 3 2 1
4 3 2 1 5 3
32
4 2 1
x x x xx
x
11-7: The Binomial Theorem
51. Expand 5 2 using the Binomial Theorem.x
5 4 3 25
3125 625 2 125 4 25 8
1 1 2
5 5 4 5 4 3
5 4 3 2 5 4 3 2
5 2
5 16 1
1 3 2 1
4 3 2 1 5 3
32
4 2 1
x x x xx
x
Step 3: Reduce as much as possible.2 2
Return to the previous slide
Continue to the next slide to see the rest of the explanation.
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11-7: The Binomial Theorem
51. Expand 5 2 using the Binomial Theorem.x
5 5 4 3 25 2 3125 6250 5000 2000 400 32x x x x x x
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Step 4: Simplify each numerator and write the final expansion.
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11-7: The Binomial TheoremGo on to the next problem
Check Answer
42. Expand 4 3 using the Binomial Theorem.a b
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11-7: The Binomial Theorem
4 4 3 2 2 3 44 3 256 768 864 432 81a b a a b a b ab b
42. Expand 4 3 using the Binomial Theorem.a b
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Note: These problems are much easier if you learn the pattern from term to term rather than try to memorize the Binomial Theorem as a formula.
Step 1: Write out the expansion. Be sure to include the negative sign.
Note the patterns: 1) the powers on the first term (4a) decrease (4, 3, 2, 1, 0)2) The powers on the second term (-3b) increase (0,1, 2, 3, 4)3) The factorial increases (0!, 1!, 2!, 3!, 4!)4) The previous power of (4a) moves to the front as a multiplier in the next term.
4 3 1 2
3
42
1 4
4 4 3 4 34 3
4 3
4
0! 1! 2!
3
4 3
4 3 2 4 2
! 4!
23 1
a a b a ba b
a b
Continue to the next slide to see the rest of the explanation.
11-7: The Binomial Theorem
42. Expand 4 3 using the Binomial Theorem.a b
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Step 2: Simplify the terms in the expansion. Remember to follow the order of operations.
11-7: The Binomial Theorem
Step 3: Reduce as much as possible.2
2
Return to the previous slide
Continue to the next slide to see the rest of the explanation.
42. Expand 4 3 using the Binomial Theorem.a b
4 3 2 24
3
1 1 2 1
3 2
256 64 3 164 4 3
4 3 2 4 3
94 3
4 27 16
1 4 1
2
2
1
3
a a b a ba b
a b
4 3 2 24
3
1 1 2 1
3 2
256 64 3 164 4 3
4 3 2 4 3
94 3
4 27 16
1 4 1
2
2
1
3
a a b a ba b
a b
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11-7: The Binomial TheoremReturn to the previous slide
Step 4: Simplify each numerator and write the final expansion.
42. Expand 4 3 using the Binomial Theorem.a b
4 4 3 2 2 3 44 3 256 768 864 432 81a b a a b a b ab b
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11-7: The Binomial TheoremGo on to the next problem
Check Answer
7
3. Use the Binomial Theorem (factorial formula) to find the fifth term in the
expansion of 3 5 .x
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Click here if you would you like to see a detailed explanation.
11-7: The Binomial Theorem
3590,625x
7
3. Use the Binomial Theorem (factorial formula) to find the fifth term in the
expansion of 3 5 .x
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Step 1: Recall the factorial formula.
!Given
! !n k knn
na
k kb ba
Continue to the next slide to see the rest of the explanation.
11-7: The Binomial Theorem
7
3. Use the Binomial Theorem (factorial formula) to find the fifth term in the
expansion of 3 5 .x
Step 2: Identify the values for n, k, a, and b.
is the power on the binomial:
is one less than the location of the term: 5 1
a 3
k
n 7
b
is the first term in the binomial:
is the second term in the binom 5
4
ial:
x
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Step 3: Plug the values into the formula.
47 4!
! !73
7
4 45x
11-7: The Binomial Theorem
7
3. Use the Binomial Theorem (factorial formula) to find the fifth term in the
expansion of 3 5 .x
Step 4: Simplify.
3 4 37!3 5 35 27 625
3! 4!x x
Step 5: State the answer.
3590,625x
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11-7: The Binomial TheoremGo on to the next problem
Check Answer
8
4. Use the Binomial Theorem (factorial formula) to find the fourth term in the
expansion of 2 9 .x y
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Click here if you would you like to see a detailed explanation.
11-7: The Binomial Theorem
5 31,306,368x y
8
4. Use the Binomial Theorem (factorial formula) to find the fourth term in the
expansion of 2 9 .x y
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Step 1: Recall the factorial formula.
!Given
! !n k knn
na
k kb ba
Continue to the next slide to see the rest of the explanation.
11-7: The Binomial Theorem
Step 2: Identify the values for n, k, a, and b.
k 3
is the power on the binomial:
is one less than the location of the term: 4 1
is the first term in the binomial:
is the second term in the binomial:
a 2
b 9
n
8
x
y
8
4. Use the Binomial Theorem (factorial formula) to find the fourth term in the
expansion of 2 9 .x y
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Step 3: Plug the values into the formula.
8 3 3
23
8
!3
!
!89x y
11-7: The Binomial Theorem
Step 4: Simplify.
5 3 5 38!2 9 56 32 729
5! 3!x y x y
Step 5: State the answer.
Return to the previous slide
8
4. Use the Binomial Theorem (factorial formula) to find the fourth term in the
expansion of 2 9 .x y
5 31,306,368x yReturn to
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Go on to the next problem
11-8: Mathematical InductionGo on to the next problem
Check Answer
1. Use the method of Mathematical Induction to prove that the statement
11 17 23 6 5 3 8 is true for all positive integers.n n n
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Click here if you would you like to see a detailed explanation.
11-8: Mathematical Induction
1. Use the method of Mathematical Induction to prove that the statement
11 17 23 6 5 3 8 is true for all positive integers.n n n
1
1
2
2
1
1. Identify , ,
2. Show the statement is true for n = S 3 8 1 11 11
3. Assume that 11 17 23 6 5 3 8
4. Proof: 11 17 23 6 5 6 1 5 3 8 6 1 5
3 8 6 6 5
3 1
1 1
4
1:n na a and S
a
k k k
k k k k k
k k k
k k
1
n
11
1 3 11
1 3 1 8
Thus S is valid.k
k k
k k
S
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Step 1: Identify a1, an, and Sn. (This helps you begin organizing your proof.)
Continue to the next slide to see the rest of the explanation.
Step 2: Let n = 1 and show that the sum formula, Sn, will be equal to the first term, a1. (This shows the statement is true for the first value of n.)
11-8: Mathematical Induction
1. Use the method of Mathematical Induction to prove that the statement
17 23 is true for all positive i6 5 nt1 e3 g s1 8 er .n nn
1 is the first term:
is the nth term formula:
is the sum formula which the statement is equal 3
6 5
to:
1
8
1
n
n
a
a n
S n n
1 1If , then n = 1 S 3 8 1 11 1 1 11 a
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Step 3: Rewrite the problem with k instead of n. (This change of variables indicates the proof is moving to the next stage. Don’t forget to begin with the word “Assume.”)
Step 4a: Find the ak+1 term. To do this change the k to (k + 1) in the an term. (This step is not part of the proof and should be done on the side.)
Return to the previous slide
11-8: Mathematical Induction
1. Use the method of Mathematical Induction to prove that the statement
11 17 23 6 5 3 8 is true for all positive integers.n n n
Assume that 11 17 23 6 5 3 8k k k
1 16 5 6 5k k ka ak
Continue to the next slide to see the rest of the explanation.
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Step 4: Add ak+1 to both sides of the statement in step 3. (Don’t forget to begin with the word “Prove or Proof.”)
Step 5: Find Sk+1. To do this change the k to (k + 1) in the Sn formula. (This step is not part of the proof and should be done on the side.) Note: This step gives you the end goal during the rest of the proof.
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11-8: Mathematical Induction
1. Use the method of Mathematical Induction to prove that the statement
11 17 23 6 5 3 8 is true for all positive integers.n n n
6 1 5Proof: 11 17 23 6 5 3 18 6 5k kk k k
13 8 3 8 or 1 3 11 1 1k kS S kk kk k k
Continue to the next slide to see the rest of the explanation.
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Step 6: Use algebra to change the right side of the statement into Sk+1. Be sure to clearly show all of the steps.
Return to the previous slide
11-8: Mathematical Induction
1. Use the method of Mathematical Induction to prove that the statement
11 17 23 6 5 3 8 is true for all positive integers.n n n
2
2
1
Proof: 11 17 23 6 5 3 8
3 8 6 6 5
3 14 11
1 3 11
6 1 5
1 3
1
8
6 5
1
k
k k k
k k k
k k
k k
k
S
k
k
k
Step 7: Make the concluding statement.
Thus Sn is valid.
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Applications/Word ProblemsGo on to the next problem
Check Answer
1. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows, how many
bricks are in the last row?
Return to
Menu
Click here if you would you like to see a detailed explanation.
Applications/Word Problems
1. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows, how many
bricks are in the last row?
There are 8 bricks in the last row.
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Step 1: Determine whether the sequence is arithmetic or geometric and write out the first few terms.
2
1
The sequence is , because it involves subtraction.
The number of bricks in the (the longest row): 36
The number of bricks in the (2 less bricks)se : cond
arit
36
first r
hme
2 34
Th
t
e
o
ic
w
n
ro
r
w
umbe
a
a
3of bricks in the (2 less brickthir s): 3d row 4 2 32a
Continue to the next slide to see the rest of the explanation.
Step 2: Determine whether you are finding a term or a sum.
Based on the first few terms in Step 1, the number of bricks in the last
row would be the last term, not a sum.
Therefore, you need to find not .n na S
Applications/Word Problems
1. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows, how many
bricks are in the last row?
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Step 3: Find an.
15
n 1
15
2 2 l
Here , because there are of bricks, and
, because each row has thess bricks an th
n = 15 1
e previous row.
Recall: a 1
So, a 36 1
36 28 8
5 r
5 2
ows
1
a n d
a
d
Step 4: Write the concluding sentence.
There are 8 bricks in the last row.
Applications/Word Problems
1. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows, how many
bricks are in the last row?
Return to the previous slide
Return to
Menu
Go on to the next problem
Applications/Word ProblemsGo on to the next problem
Check Answer
2. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows of bricks,
how many bricks were used to make the patio?
Return to
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Click here if you would you like to see a detailed explanation.
Applications/Word Problems
To make the patio 330 bricks were used.
2. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows of bricks,
how many bricks were used to make the patio?
Return to
Menu
Go on to the next problem
Step 1: Determine whether the sequence is arithmetic or geometric and write out the first few terms.
2
1
The sequence is , because it involves subtraction.
The number of bricks in the (the longest row): 36
The number of bricks in the (2 less bricks)se : cond
arit
36
first r
hme
2 34
Th
t
e
o
ic
w
n
ro
r
w
umbe
a
a
3of bricks in the (2 less brickthir s): 3d row 4 2 32a
Continue to the next slide to see the rest of the explanation.
Step 2: Determine whether you are finding a term or a sum.
Since each term gives only the number of bricks in that row,
the number of bricks used in the entire patio would be the sum, not a term.
Therefore, you need to find not .n nS a
Applications/Word Problems
2. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows of bricks,
how many bricks were used to make the patio?
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Go on to the next problem
Step 3: Find an.
1
n
Recall: 2
Here , because there are of bricks, but the value of
a (the number of bricks in the last row) isn't known and must be found first.
Now, , becaus
n
e 2 2 leea
= 15 15
ch
ro
row has
w
s
s
n nn
S a a
d
n
15
n 1
15
than the previous row.
So, a is found as follows:
Recall: a 1
Plug in the values,
s bricks
1a 36 15
36 8 8
2
2
a n d
a
Applications/Word ProblemsReturn to the previous slide
2. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows of bricks,
how many bricks were used to make the patio?
Continue to the next slide to see the rest of the explanation.
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Go on to the next problem
Step 4: Find Sn.
1
15
Now, 2
15 15Plug in the known values: 36 8 44 15 22 330
2 2
n nn
S a a
S
Step 5: Write the concluding sentence.
Applications/Word ProblemsReturn to the previous slide
2. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. If there are 15 rows of bricks,
how many bricks were used to make the patio?
To make the patio, 330 bricks were used.
Return to
Menu
Go on to the next problem
Applications/Word ProblemsGo on to the next problem
Check Answer
3. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. How many bricks are in the tenth row?
Return to
Menu
Click here if you would you like to see a detailed explanation.
Applications/Word Problems
There are 18 bricks in the tenth row.
3. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. How many bricks are in the tenth row?
Return to
Menu
Go on to the next problem
Step 1: Determine whether the sequence is arithmetic or geometric and write out the first few terms.
2
1
The sequence is , because it involves subtraction.
The number of bricks in the (the longest row): 36
The number of bricks in the (2 less bricks)se : cond
arit
36
first r
hme
2 34
Th
t
e
o
ic
w
n
ro
r
w
umbe
a
a
3of bricks in the (2 less brickthir s): 3d row 4 2 32a
Continue to the next slide to see the rest of the explanation.
Step 2: Determine whether you are finding a term or a sum.
10
Since each term gives the number of bricks in that row,
the number of bricks in the tenth row would be the tenth term, a .
Applications/Word Problems
3. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. How many bricks are in the tenth row?
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Go on to the next problem
Step 3: Find an.
10
10
n 1
10
Here , for the , and , because each row has
than the previous row
2
.
So, a is found as follows:
Recall: a 1
Plug in the values, a 36 1
36 18 18
n = 10 tenth row
1
2
0
less bricks
2
a n d
a
d
Applications/Word ProblemsReturn to the previous slide
There are 18 bricks in the tenth row.
3. A patio is in the shape of a triangle. The longest row has 36 bricks.
Each row after that has two less bricks. How many bricks are in the tenth row?
Step 4: Write the concluding sentence.
Return to
Menu
Go on to the next problem
Applications/Word ProblemsGo on to the next problem
Check Answer
4. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50%
as far into the ground. How far in the ground will the post be after 6 strikes?
Return to
Menu
Click here if you would you like to see a detailed explanation.
Applications/Word Problems
4. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50%
as far into the ground. How far in the ground will the post be after 6 strikes?
5The post will be 23 inches in the ground after 6 strikes.
8
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Go on to the next problem
Step 1: Determine whether the sequence is arithmetic or geometric and write out the first few terms.
2
1
3
The sequence is , because it involves multiplication.
The depth with the : 12 inches
The depth with the : 12 0.5 6 inches
The dep
second
th wit
geome
h the
tri
t
first st
hird st
st
rik
r
e
r
:
i
ik
c
.5
ke
e
6 0
a
a
a 3 inches
Continue to the next slide to see the rest of the explanation.
Step 2: Determine whether the problem is asking you to find a term or a sum.
Applications/Word Problems
4. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50%
as far into the ground. How far in the ground will the post be after 6 strikes?
The wording “after 6 strikes” means we are looking for the total, or sum, of the first 6 depths, S6. (Note: “on the 6th strike” would have been a6.)
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Step 3: Find Sn.
1
6
6
6
1n
Here , for the , and .
So, S is found as
n = 6 after 0.5 (
follows:
Recall: S1
5Plug in the values, S
s forix
12
strik 5
23
0%)
0.5
es
12
51 80.
na a r
r
r
Applications/Word ProblemsReturn to the previous slide
Step 4: Write the concluding sentence.
4. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50%
as far into the ground. How far in the ground will the post be after 6 strikes?
5The post will be 23 inches in the ground after 6 strikes.
8Return
to Menu
Go on to the next problem
Applications/Word ProblemsGo on to the next problem
Check Answer
5. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50%
as far into the ground. How far will the post go into the ground on the 4th strike?
Return to
Menu
Click here if you would you like to see a detailed explanation.
Applications/Word Problems
1The post will go 1 inches into the ground on the 4th strike.
2
5. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50%
as far into the ground. How far will the post go into the ground on the 4th strike?
Return to
Menu
Go on to the next problem
Step 1: Determine whether the sequence is arithmetic or geometric and write out the first few terms.
2
1
3
The sequence is , because it involves multiplication.
The depth with the : 12 inches
The depth with the : 12 0.5 6 inches
The dep
second
th wit
geome
h the
tri
t
first st
hird st
st
rik
r
e
r
:
i
ik
c
.5
ke
e
6 0
a
a
a 3 inches
Continue to the next slide to see the rest of the explanation.
Step 2: Determine whether the problem is asking you to find a term or a sum.
Applications/Word Problems
The wording “on the 4th strike” means we are looking for fourth term, a4. (Note: “after 4 strikes” would have been S4.)
5. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50%
as far into the ground. How far will the post go into the ground on the 4th strike?
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Go on to the next problem
Step 3: Find an.
1n 1
4
414
Here , for the , and .
So, a is found as follows:
Recall: a
1Plug
n = 4 on the fourth s
in the values, a
0.5 (for 50%)
0.5
tri
12 12
ke
na r
r
Applications/Word ProblemsReturn to the previous slide
Step 4: Write the concluding sentence.
5. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50%
as far into the ground. How far will the post go into the ground on the 4th strike?
1The post will go 1 inches into the ground on the 4th strike.
2Return
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Go on to the next problem
Applications/Word Problems
Check Answer
6. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50% as far
into the ground. What is the total distance the post can be driven into the ground?
Return to
Menu
Click here if you would you like to see a detailed explanation.
Applications/Word Problems
The total distance the post can be driven into the ground is 24 inches.
6. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50% as far
into the ground. What is the total distance the post can be driven into the ground?
Return to
Menu
Step 1: Determine whether the sequence is arithmetic or geometric and write out the first few terms.
2
1
3
The sequence is , because it involves multiplication.
The depth with the : 12 inches
The depth with the : 12 0.5 6 inches
The dep
second
th wit
geome
h the
tri
t
first st
hird st
st
rik
r
e
r
:
i
ik
c
.5
ke
e
6 0
a
a
a 3 inches
Continue to the next slide to see the rest of the explanation.
Step 2: Determine whether the problem is asking you to find a term or a sum.
Applications/Word Problems
The word “total” implies we need to find a sum. The absence of a number of strikes to stop at means this is an “infinite geometric series” problem.
6. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50% as far
into the ground. What is the total distance the post can be driven into the ground?
Return to
Menu
Step 3: Find S.
1
Here there is and .
So, is found as follows:
Recall: , where 1 11
12Since 1
no 0.5 (for 50%)
0. 1,
valu
241
S
S
50.5
e r
ar
S
n
r
Applications/Word ProblemsReturn to the previous slide
Step 4: Write the concluding sentence.
6. A machine is driving a post into the ground. On the first strike the post goes
12 inches into the ground. Each strike after the first the post goes only 50% as far
into the ground. What is the total distance the post can be driven into the ground?
The total distance the post can be driven into the ground is 24 inches.
Return to
Menu