Competition problem — Algebra — Maths Olympiad Prep

Library / /19 of 100

Algebra Difficulty 4.2 AIME Find the answer China

Let $\{a_n\}$ be an arithmetic progression with common difference $d$ ( $d \neq 0$ ) and $\{b_n\}$ be a geometric progression with common ratio $q$ , where $q$ is a positive rational number less than $1$ . If $a_1 = d$ , $b_1 = d^2$ and $\frac{a_1^2 + a_2^2 + a_3^2}{b_1 + b_2 + b_3}$ is a positive integer, then $q$ equals ______.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

As
$\begin{aligned} \frac{a_1^2 + a_2^2 + a_3^2}{b_1 + b_2 + b_3} &= \frac{a_1^2 + (a_1+d)^2 + (a_1+2d)^2}{b_1 + b_1q + b_1q^2} \\ &= \frac{14}{1+q+q^2} = m \end{aligned}$
is a positive integer, we get $1 + q + q^2 = \frac{14}{m}$ . Then
$q = -\frac{1}{2} + \sqrt{\frac{1}{4} + \frac{14}{m} - 1} = -\frac{1}{2} + \sqrt{\frac{56-3m}{4m}}$
Since $q$ is a positive rational number less than $1$ , we have $1 < \frac{14}{m} < 3$ , i.e. $5 \le m \le 13$ , and $\frac{56-3m}{4m}$ is the square of a rational number. We can verify that only $m = 8$ meets the required. That means $q = \frac{1}{2}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.