1988 AJHSME Problems/Problem 19

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 $\boxed{A}$

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
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1988 AJHSME Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

See Also