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 "1960 AHSME Problems/Problem 36"

m (→‎Problem 36)
m
 
(One intermediate revision by one other user not shown)
Line 27: Line 27:
  
 
The answer is <math>\boxed{\textbf{(B)}}</math>.
 
The answer is <math>\boxed{\textbf{(B)}}</math>.
 +
 +
 +
==Video Solution==
 +
https://youtu.be/ZdM2ou5Gsuw?t=89
 +
 +
MathProblemSolvingSkills.com
 +
  
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=35|num-a=37}}
 
{{AHSME 40p box|year=1960|num-b=35|num-a=37}}
 +
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 21:25, 28 December 2023

Problem

Let $s_1, s_2, s_3$ be the respective sums of $n, 2n, 3n$ terms of the same arithmetic progression with $a$ as the first term and $d$ as the common difference. Let $R=s_3-s_2-s_1$. Then $R$ is dependent on:

$\textbf{(A)}\ a\text{ }\text{and}\text{ }d\qquad \textbf{(B)}\ d\text{ }\text{and}\text{ }n\qquad \textbf{(C)}\ a\text{ }\text{and}\text{ }n\qquad \textbf{(D)}\ a, d,\text{ }\text{and}\text{ }n\qquad \textbf{(E)}\ \text{neither} \text{ } a \text{ } \text{nor} \text{ } d \text{ } \text{nor} \text{ } n$

Solution

The nth term of the sequence with first term $a$ is $a + d(n-1)$. That means the sum of the first $n$ terms is with first term $a$ is $\frac{n(2a + dn -d)}{2}$.

The 2nth term of the sequence with first term $a$ is $a + d(2n-1)$. That means the sum of the first $2n$ terms is with first term $a$ is $\frac{2n(2a + 2dn -d)}{2}$.

The 3nth term of the sequence with first term $a$ is $a + d(3n-1)$. That means the sum of the first $3n$ terms is with first term $a$ is $\frac{3n(2a + 3dn -d)}{2}$.

Substituting $s_1$, $s_2$, and $s_3$ with its respective values results in

\[\frac{3n(2a + 3dn -d)}{2}-\frac{2n(2a + 2dn -d)}{2}-\frac{n(2a + dn -d)}{2}\] \[\frac{6an + 9dn^2 -3nd}{2}-\frac{4an + 4dn^2 - 2nd}{2}-\frac{2an + dn^2 - dn}{2}\]

\[\frac{4dn^2}{2}\]

\[2dn^2\]

The answer is $\boxed{\textbf{(B)}}$.


Video Solution

https://youtu.be/ZdM2ou5Gsuw?t=89

MathProblemSolvingSkills.com


See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 35 		Followed by
Problem 37
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_36&oldid=208616"