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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>
  
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==Solution==
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==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>
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==Solution 2==
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Alternatively you could create an equation for the arithmetic sequence: <math>a_{n}=a_{1}+4(n-1)=1+4n-4=4n-3</math>
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<math>a_{100}=4(100)-3=397</math>, or <math>\boxed{A}</math>
  
 
==See Also==
 
==See Also==
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{{AJHSME box|year=1988|num-b=18|num-a=20}}
 
{{AJHSME box|year=1988|num-b=18|num-a=20}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 09:51, 28 June 2023

Problem

What is the $100\text{th}$ number in the arithmetic sequence: $1,5,9,13,17,21,25,...$?

$\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405$

Solution 1

To get from the $1^\text{st}$ term of an arithmetic sequence to the $100^\text{th}$ term, we must add the common difference $99$ times. The first term is $1$ and the common difference is $5-1=9-5=13-9=\cdots = 4$, so the $100^\text{th}$ term is \[1+4(99)=397 \rightarrow \boxed{\text{A}}\]

Solution 2

Alternatively you could create an equation for the arithmetic sequence: $a_{n}=a_{1}+4(n-1)=1+4n-4=4n-3$

$a_{100}=4(100)-3=397$, or $\boxed{A}$

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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