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Difference between revisions of "2025 AMC 8 Problems/Problem 4"
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Lucius is counting backward by <math>7</math>s. His first three numbers are <math>100</math>, <math>93</math>, and <math>86</math>. What is his <math>10</math>th number?
 
Lucius is counting backward by <math>7</math>s. His first three numbers are <math>100</math>, <math>93</math>, and <math>86</math>. What is his <math>10</math>th number?
  
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<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47</math>
 
==Solution==
 
==Solution==
  

Revision as of 19:27, 29 January 2025

Lucius is counting backward by $7$s. His first three numbers are $100$, $93$, and $86$. What is his $10$th number?

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47$

Solution

By the formula for the $n$th term of an arithmetic sequence, we get that the answer is $a+d(n-1)$ where $a=100, d=-7$ and $n=10$ which is $100 - 7(10 - 1) = \boxed{\text{(B)\ 37}}$.

~Soupboy0

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