Difference between revisions of "1973 AHSME Problems/Problem 7"

(Solution to Problem 7)
 
(See Also)
 
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==See Also==
 
==See Also==
{{AHSME 35p box|year=1973|num-b=6|num-a=8}}
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{{AHSME 30p box|year=1973|num-b=6|num-a=8}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 12:58, 20 February 2020

Problem

The sum of all integers between 50 and 350 which end in 1 is

$\textbf{(A)}\ 5880\qquad\textbf{(B)}\ 5539\qquad\textbf{(C)}\ 5208\qquad\textbf{(D)}\ 4877\qquad\textbf{(E)}\ 4566$

Solution

The numbers that we are adding are $51,61,71 \cdots 341$. The numbers are part of an arithmetic series with first term $51$, last term $341$, common difference $10$, and $30$ terms. Using the arithmetic series formula, the sum of the terms is $\tfrac{30 \cdot 392}{2} = \boxed{\textbf{(A)}\ 5880}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AHSME Problems and Solutions