Euclid 2020/Problem 4

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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Euclid 2020/Problem 4

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