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Difference between revisions of "1969 AHSME Problems/Problem 33"

(Created page with "== Problem == Let <math>S_n</math> and <math>T_n</math> be the respective sums of the first <math>n</math> terms of two arithmetic series. If <math>S_n:T_n=(7n+1):(4n+27)</math>...")
 
(Solution to Problem 33 - finally solved it!)
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== Solution ==
 
== Solution ==
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<math>\fbox{A}</math>
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Let <math>S</math> be the first arithmetic sequence and <math>T</math> be the second arithmetic sequence.  If <math>n = 1</math>, then <math>S_1:T_1 = 8:31</math>.  Since <math>S_1</math> and <math>T_1</math> are just the first term, the first term of <math>S</math> is <math>8a</math> and the first term of <math>T</math> is <math>31a</math> for some <math>a</math>.  If <math>n = 2</math>, then <math>S_2:T_2 = 15:35 = 3:7</math>, so the sum of the first two terms of <math>S</math> is <math>3b</math> and the sum of the first two terms of <math>T</math> is <math>7b</math> for some <math>b</math>.  Thus, the second term of <math>S</math> is <math>3b-8a</math> and the second term of <math>T</math> is <math>7b - 31a</math>, so the common difference of <math>S</math> is <math>3b-16a</math> and the common difference of <math>T</math> is <math>6b-62a</math>.
  
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== See also ==
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Thus, using the first terms and common differences, the sum of the first three terms of <math>S</math> equals <math>\tfrac{1}{2} \cdot 3(16a + 2(-16a + 3b))</math>, and the sum of the first three terms of <math>T</math> equals <math>\tfrac{1}{2} \cdot 3(62a + 2(-62a + 7b))</math>.  That means
 +
<cmath>\frac{16a + 2(-16a + 3b)}{62a + 2(-62a + 7b)} = \frac{22}{39}</cmath>
 +
<cmath>\frac{6b-16a}{14b-62a} = \frac{22}{39}</cmath>
 +
<cmath>\frac{3b-8a}{7b-31a} = \frac{22}{39}</cmath>
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<cmath>117b - 312a = 154b - 682a</cmath>
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<cmath>-37b = -370a</cmath>
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<cmath>b = 10a</cmath>
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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>.
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== See Also ==
 
{{AHSME 35p box|year=1969|num-b=32|num-a=34}}   
 
{{AHSME 35p box|year=1969|num-b=32|num-a=34}}   
  
 
[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:03, 21 June 2018

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 $6b-62a$.

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 $148a:111a = \boxed{textbf{(A) } 4:3}$.

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

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