Difference between revisions of "1988 AJHSME Problems/Problem 19"
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<math>\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405</math> | <math>\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405</math> | ||
− | ==Solution== | + | ==Solution 1== |
To get from the <math>1^\text{st}</math> term of an arithmetic sequence to the <math>100^\text{th}</math> term, we must add the common [[difference]] <math>99</math> times. The first term is <math>1</math> and the common difference is <math>5-1=9-5=13-9=\cdots = 4</math>, so the <math>100^\text{th}</math> term is <cmath>1+4(99)=397 \rightarrow \boxed{\text{A}}</cmath> | To get from the <math>1^\text{st}</math> term of an arithmetic sequence to the <math>100^\text{th}</math> term, we must add the common [[difference]] <math>99</math> times. The first term is <math>1</math> and the common difference is <math>5-1=9-5=13-9=\cdots = 4</math>, so the <math>100^\text{th}</math> term is <cmath>1+4(99)=397 \rightarrow \boxed{\text{A}}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Alternatively you could create an equation for the arithmetic sequence: <math>a_{n}=a_{1}+4(n-1)=1+4n-4=4n-3</math> | ||
+ | |||
+ | <math>a_{100}=4(100)-3=397</math>, or <math>\boxed{A}</math> | ||
==See Also== | ==See Also== |
Latest revision as of 09:51, 28 June 2023
Contents
Problem
What is the number in the arithmetic sequence: ?
Solution 1
To get from the term of an arithmetic sequence to the term, we must add the common difference times. The first term is and the common difference is , so the term is
Solution 2
Alternatively you could create an equation for the arithmetic sequence:
, or
See Also
1988 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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