Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1965 AHSME Problems/Problem 20"

(Created page with "In order to calculate the <math>r</math>th term of this arithmetic sequence, we can subtract the sum of the first <math>r-1</math> terms from the sum of the first <math>r</mat...")
 
(see also box, put letter in answer box)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
−
In order to calculate the <math>r</math>th term of this arithmetic sequence, we can subtract the sum of the first <math>r-1</math> terms from the sum of the first <math>r</math> terms of the sequence. Plugging in <math>r</math> and <math>r-1</math> as values of <math>n</math> in the given expression and subtracting yields <cmath>(3r^2+2r)-(3r^2-4r+1)</cmath>. Simplifying gives us the final answer of <math>\boxed{6r-1}</math>.
+
== Problem ==
 +
 
 +
For every <math>n</math> the sum of n terms of an arithmetic progression is <math>2n + 3n^2</math>. The <math>r</math>th term is:
 +
 
 +
<math>\textbf{(A)}\ 3r^2 \qquad \textbf{(B) }\ 3r^2 + 2r \qquad \textbf{(C) }\ 6r - 1 \qquad \textbf{(D) }\ 5r + 5 \qquad \textbf{(E) }\ 6r+2\qquad </math>
 +
 
 +
== Solution ==
 +
 
 +
In order to calculate the <math>r</math>th term of this arithmetic sequence, we can subtract the sum of the first <math>r-1</math> terms from the sum of the first <math>r</math> terms of the sequence. Plugging in <math>r</math> and <math>r-1</math> as values of <math>n</math> in the given expression and subtracting yields <cmath>(3r^2+2r)-(3r^2-4r+1).</cmath> Simplifying gives us the final answer of <math>\boxed{\textbf{(C) }6r-1}</math>.
 +
 
 +
== See Also ==
 +
{{AHSME 40p box|year=1965|num-b=19|num-a=21}}
 +
{{MAA Notice}}

Latest revision as of 16:13, 18 July 2024

Problem

For every $n$ the sum of n terms of an arithmetic progression is $2n + 3n^2$. The $r$th term is:

$\textbf{(A)}\ 3r^2 \qquad \textbf{(B) }\ 3r^2 + 2r \qquad \textbf{(C) }\ 6r - 1 \qquad \textbf{(D) }\ 5r + 5 \qquad \textbf{(E) }\ 6r+2\qquad$

Solution

In order to calculate the $r$th term of this arithmetic sequence, we can subtract the sum of the first $r-1$ terms from the sum of the first $r$ terms of the sequence. Plugging in $r$ and $r-1$ as values of $n$ in the given expression and subtracting yields \[(3r^2+2r)-(3r^2-4r+1).\] Simplifying gives us the final answer of $\boxed{\textbf{(C) }6r-1}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1965_AHSME_Problems/Problem_20&oldid=223034"