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 "1969 AHSME Problems/Problem 33"

m (Fixed Typo)
Line 20: Line 20:
 
<cmath>b = 10a</cmath>
 
<cmath>b = 10a</cmath>
 
With the substitution, the common difference of <math>S</math> is <math>14a</math>, and the common difference of <math>T</math> is <math>8a</math>.  That means the <math>11^\text{th}</math> term of <math>S</math> is <math>8a + 10(14a) = 148a</math>, and the <math>11^\text{th}</math> term of <math>T</math> is <math>31a + 10(8a) = 111a</math>.  Thus, the ratio of the eleventh term of the first series to the eleventh term of the second series is <math>148a:111a = \boxed{\textbf{(A) } 4:3}</math>.
 
With the substitution, the common difference of <math>S</math> is <math>14a</math>, and the common difference of <math>T</math> is <math>8a</math>.  That means the <math>11^\text{th}</math> term of <math>S</math> is <math>8a + 10(14a) = 148a</math>, and the <math>11^\text{th}</math> term of <math>T</math> is <math>31a + 10(8a) = 111a</math>.  Thus, the ratio of the eleventh term of the first series to the eleventh term of the second series is <math>148a:111a = \boxed{\textbf{(A) } 4:3}</math>.
 +
 +
== Solution 2 (Quick Sol) ==
 +
Note that the sum of arithmetic sequences can be expressed as a quadratic polynomial of <math>n</math>. From the ratio given in the problem, multiply <math>n</math> to both sides to get quadratic polynomials. <cmath>S_n = 7n^2+n, T_n = 4n^2+27n</cmath>. From there, the 11th term for <math>S_n</math> and <math>T_n</math> can becalculated. <cmath>S_11 - S_10 = 7*11^2+11 - (7*10^2+10) = 148</cmath> <cmath>T_11 - T_10 = 4*11^2+27*11 - (4*10^2+27*10) = 111</cmath>. The ratio is <math>148 : 111 = \boxed{\textbf{(A) } 4:3}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 00:50, 1 June 2021

Problem

Let $S_n$ and $T_n$ be the respective sums of the first $n$ terms of two arithmetic series. If $S_n:T_n=(7n+1):(4n+27)$ for all $n$, the ratio of the eleventh term of the first series to the eleventh term of the second series is:

$\text{(A) } 4:3\quad \text{(B) } 3:2\quad \text{(C) } 7:4\quad \text{(D) } 78:71\quad \text{(E) undetermined}$

Solution

Let $S$ be the first arithmetic sequence and $T$ be the second arithmetic sequence. If $n = 1$, then $S_1:T_1 = 8:31$. Since $S_1$ and $T_1$ are just the first term, the first term of $S$ is $8a$ and the first term of $T$ is $31a$ for some $a$. If $n = 2$, then $S_2:T_2 = 15:35 = 3:7$, so the sum of the first two terms of $S$ is $3b$ and the sum of the first two terms of $T$ is $7b$ for some $b$. Thus, the second term of $S$ is $3b-8a$ and the second term of $T$ is $7b - 31a$, so the common difference of $S$ is $3b-16a$ and the common difference of $T$ is $7b-62a$.

Thus, using the first terms and common differences, the sum of the first three terms of $S$ equals $\tfrac{1}{2} \cdot 3(16a + 2(-16a + 3b))$, and the sum of the first three terms of $T$ equals $\tfrac{1}{2} \cdot 3(62a + 2(-62a + 7b))$. That means \[\frac{16a + 2(-16a + 3b)}{62a + 2(-62a + 7b)} = \frac{22}{39}\] \[\frac{6b-16a}{14b-62a} = \frac{22}{39}\] \[\frac{3b-8a}{7b-31a} = \frac{22}{39}\] \[117b - 312a = 154b - 682a\] \[-37b = -370a\] \[b = 10a\] With the substitution, the common difference of $S$ is $14a$, and the common difference of $T$ is $8a$. That means the $11^\text{th}$ term of $S$ is $8a + 10(14a) = 148a$, and the $11^\text{th}$ term of $T$ is $31a + 10(8a) = 111a$. Thus, the ratio of the eleventh term of the first series to the eleventh term of the second series is $148a:111a = \boxed{\textbf{(A) } 4:3}$.

Solution 2 (Quick Sol)

Note that the sum of arithmetic sequences can be expressed as a quadratic polynomial of $n$. From the ratio given in the problem, multiply $n$ to both sides to get quadratic polynomials. \[S_n = 7n^2+n, T_n = 4n^2+27n\]. From there, the 11th term for $S_n$ and $T_n$ can becalculated. \[S_11 - S_10 = 7*11^2+11 - (7*10^2+10) = 148\] \[T_11 - T_10 = 4*11^2+27*11 - (4*10^2+27*10) = 111\]. The ratio is $148 : 111 = \boxed{\textbf{(A) } 4:3}$.

See Also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
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
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=1969_AHSME_Problems/Problem_33&oldid=154655"