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Difference between revisions of "Euclid 2020/Problem 4"
 
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==Solution==
 
==Solution==
 
(a) Because both <math>\frac{5}{9}</math> and <math>\frac47</math> are less than one, we have <math>a<b</math>. Because the difference between <math>a</math> and <math>b</math> is 15, we have <math>a=b-15</math>, so we have <math>\frac59<\frac{b-15}b<\frac47</math>. Using the left side of the equation, we have <math>\frac59<\frac{b-15}b</math> so <math>5b<9b-135</math>, so <math>135<4b</math>, so <math>b>\frac{135}4</math>. Using the right side of the equation, we have <math>\frac{b-15}b<\frac47</math>, so <math>7b-105<4b</math>, so <math>3b<105</math>, so <math>b<35</math>. Combining both inequalities will give us <math>35>b>\frac{135}{4}</math>, so the only possible integer <math>b</math> that satisfies this inequality is <math>b=34</math>. Therefore, our answer is <math>\boxed{\frac{19}{34}}</math>.
 
(a) Because both <math>\frac{5}{9}</math> and <math>\frac47</math> are less than one, we have <math>a<b</math>. Because the difference between <math>a</math> and <math>b</math> is 15, we have <math>a=b-15</math>, so we have <math>\frac59<\frac{b-15}b<\frac47</math>. Using the left side of the equation, we have <math>\frac59<\frac{b-15}b</math> so <math>5b<9b-135</math>, so <math>135<4b</math>, so <math>b>\frac{135}4</math>. Using the right side of the equation, we have <math>\frac{b-15}b<\frac47</math>, so <math>7b-105<4b</math>, so <math>3b<105</math>, so <math>b<35</math>. Combining both inequalities will give us <math>35>b>\frac{135}{4}</math>, so the only possible integer <math>b</math> that satisfies this inequality is <math>b=34</math>. Therefore, our answer is <math>\boxed{\frac{19}{34}}</math>.
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(b) By bashing it out, we have the 4th term of the geometric sequence be <math>\frac{5}{4}</math> and the 6th term be <math>\frac{5}{16}</math>. Because the arithmetic sequence has first term 10 and common difference <math>d</math>, its 4th term will be <math>10+3d</math> and its 6th term will be <math>10+5d</math>. Our equation will then be <math>\frac{\frac{5}{16}}{\frac{5}{4}}=\frac{10+3d}{10+5d}</math>, so <math>\frac14=\frac{10+3d}{10+5d}</math>, so <math>10+5d=40+12d</math>, so <math>-30=7d</math>, so <math>d=\frac{-30}{7}</math>. Therefore, our answer is <math>\boxed{\frac{-30}{7}}</math>.
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~Yuhao2012

Latest revision as of 19:00, 12 October 2025

Problem

(a) The positive integers $a$ and $b$ have no common divisor larger than $1$. If the difference between b and a is $15$ and $\frac59<\frac{a}b<\frac47$, what is the value of $\frac{a}{b}$?

(b) A geometric sequence has  first term $10$ and common ratio $1/2$ . An arithmetic sequence has  first term $10$ and common difference $d$. The ratio of the $6th$ term in the geometric sequence to the $4th$ term in the geometric sequence equals the ratio of the $6th$ term in the arithmetic sequence to the $4th$ term in the arithmetic sequence. Determine all possible values of $d$. (An arithmetic sequence is a sequence in which each term after the  first is obtained from the previous term by adding a constant, called the common difference. For example, $3; 5; 7; 9$ are the  first four terms of an arithmetic sequence. A geometric sequence is a sequence in which each term after the  first is obtained from the previous term by multiplying it by a non-zero constant, called the common ratio. For example, $3, 6, 12$ is a geometric sequence with three terms.)

Solution

(a) Because both $\frac{5}{9}$ and $\frac47$ are less than one, we have $a<b$. Because the difference between $a$ and $b$ is 15, we have $a=b-15$, so we have $\frac59<\frac{b-15}b<\frac47$. Using the left side of the equation, we have $\frac59<\frac{b-15}b$ so $5b<9b-135$, so $135<4b$, so $b>\frac{135}4$. Using the right side of the equation, we have $\frac{b-15}b<\frac47$, so $7b-105<4b$, so $3b<105$, so $b<35$. Combining both inequalities will give us $35>b>\frac{135}{4}$, so the only possible integer $b$ that satisfies this inequality is $b=34$. Therefore, our answer is $\boxed{\frac{19}{34}}$.

(b) By bashing it out, we have the 4th term of the geometric sequence be $\frac{5}{4}$ and the 6th term be $\frac{5}{16}$. Because the arithmetic sequence has first term 10 and common difference $d$, its 4th term will be $10+3d$ and its 6th term will be $10+5d$. Our equation will then be $\frac{\frac{5}{16}}{\frac{5}{4}}=\frac{10+3d}{10+5d}$, so $\frac14=\frac{10+3d}{10+5d}$, so $10+5d=40+12d$, so $-30=7d$, so $d=\frac{-30}{7}$. Therefore, our answer is $\boxed{\frac{-30}{7}}$.

~Yuhao2012

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