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2025 AMC 8 Problems/Problem 4: Difference between revisions

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The 2025 AMC 8 is not held yet. '''Please do not post false problems.'''
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?
 
 
==Solution==
 
By the formula for the <math>n</math>th term of an arithmetic sequence, we get that the answer is <math>a+d(n-1)</math> where <math>a=100, d=-7</math> and <math>n=10</math> which is <math>100 - 7(10 - 1) = \boxed{\text{(B)\ 37}}</math>.
 
~Soupboy0

Revision as of 19:24, 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?


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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