Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Find the answer China

It is known that $\{a_n\}$ is an arithmetic sequence with non-zero common difference and $\{b_n\}$ a geometric sequence, satisfying $a_1 = 3$ , $b_1 = 1$ , $a_2 = b_2$ , $3a_5 = b_3$ ; furthermore, there are constants $\alpha$ and $\beta$ such that for every positive integer $n$ , we have $a_n = \log_a b_n + \beta$ . Then $\alpha + \beta = \underline{\hspace{2cm}}$ .

Your answer

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

Solution

Let the common difference of $\{a_n\}$ be $d$ and the common ratio of $\{b_n\}$ be $q$ . Then
$3 + d = q, \qquad \textcircled{1}$
$3(3 + 4d) = q^2. \qquad \textcircled{2}$
Substituting $\textcircled{1}$ into $\textcircled{2}$ , we have $9 + 12d = d^2 + 6d + 9$ . Then we get $d = 6$ and $q = 9$ .
Therefore, $3 + 6(n - 1) = \log_a 9^{n-1} + \beta$ or $6n - 3 = (n - 1)\log_a 9 + \beta$ holds for every positive integer $n$ . Letting $n = 1$ and $n = 2$ in turn, we find that $\alpha = \sqrt[3]{3}$ and $\beta = 3$ .
Consequently, $\alpha + \beta = \sqrt[3]{3} + 3$ . $\square$

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