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Problem 414

AMC 12 late, AIME early

Algebra Difficulty 4.8 Find the answer

12. Let the positive term geometric sequence $\left\{a_{n}\right\}$ satisfy $a_{7}=a_{6}+2 a_{5}$ . If there exist two terms $a_{n} 、 a_{m}$ such that $\sqrt{a_{m} a_{n}}=4 a_{1}$ , then the minimum value of $\frac{1}{m}+\frac{4}{n}$ is $(\quad)$ .
(A) $\frac{25}{3}$
(B) $\frac{25}{6}$
(C) $\frac{5}{3}$
(D) $\frac{3}{2}$

Your answer

Official solution

12. D.

Let the common ratio of the sequence be $q(q>0)$ .
From the problem, we know $q^{2}=q+2 \Rightarrow q=2$ .
Also, $\sqrt{a_{m} a_{n}}=4 a_{1}$ , then
$\begin{array}{l} q^{m+n-2}=4^{2}=2^{4} \\ \Rightarrow m+n-2=4 \Rightarrow m+n=6 . \end{array}$

Notice,
$\begin{array}{l} \frac{1}{m}+\frac{4}{n}=\left(\frac{1}{m}+\frac{4}{n}\right) \frac{m+n}{6} \\ =\frac{1}{6}\left(5+\frac{n}{m}+\frac{4 m}{n}\right) \\ \geqslant \frac{1}{6}(5+2 \sqrt{4})=\frac{3}{2} . \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.