Competition problem — Algebra — Maths Olympiad Prep

Library / /118 of 520

Algebra Difficulty 2.9 Junior Find the answer

Given an arithmetic sequence $\{a_n\}$ with the first term $a_1=1$ and a common difference $d>0$ , and a geometric sequence $\{b_n\}$ , satisfying $b_2=a_2$ , $b_3=a_5$ , $b_4=a_{14}$ .
(1) Find the general term of the sequences $\{a_n\}$ and $\{b_n\}$ .
(2) Let the sequence $\{c_n\}$ satisfy $c_n=2a_n-18$ , find the minimum value of the sum of the first $n$ terms $S_n$ of the sequence $\{c_n\}$ , and the value of $n$ at this minimum.

Your answer

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

Solution

(1) From the given conditions, we have $(1+4d)^2=(1+d)(1+13d)$ , and since $d>0$ , solving this gives $d=2$ .
Therefore, the general term of $\{a_n\}$ is $a_n=2n-1$ .
Since $b_2=a_2=3$ , $b_3=a_5=9$ ,
the common ratio of $\{b_n\}$ is 3, thus the general term of $\{b_n\}$ is $b_n=3^n-1$ .
(2) Since $c_n=2a_n-18=4n-20$ ,
setting $c_n\leq 0$ gives $n\leq 5$ .
Therefore, when $n=4$ or $n=5$ , $S_n$ reaches its minimum value of $\boxed{-40}$ .

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