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title: A Quotient Sequence from Arithmetic and Geometric Progressions
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# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter
---

## Introduction
It is known that $\{a_n\}$ is an infinite arithmetic sequence with the first term being $a$ and a tolerance of 2. $\{b_n\}$ is an infinite arithmetic sequence with the first term being 1 and a common ratio of 2. Let’s call it $\lambda_n = \dfrac{a_1 + a_2 + \cdots + a_n}{b_n}$. The following four conclusions are given:
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①When $0 < a < 2$, there is $\lambda_1 < \lambda_2 < \lambda_3$;

② There exists $a \in \mathbb{R}$, so that the first 2025 items of $\{\lambda_n\}$ are monotonically increasing sequences;

③For any $a > 0$, $\{\lambda_n\}$ are monotonically decreasing sequences from the third item onwards;

④ If and only if $a = 2$, there exists $k \in \mathbb{N}^*$ such that $\lambda_k = \lambda_{k+1}$.

The serial numbers of all correct conclusions are ___ .


First, remember $S_n=\sum_{i=1}^na_i$ and calculate the first three terms of ${S_n},{b_n},{\lambda_n}$:

| n | $S_n$ | $b_n$ | $\lambda_n$ |
| --- | --- | --- | --- |
| 1 | $a$ | 1 | $a$ |
| 2 | $2a+2$ | 2 | $a+1$ |
| 3 | $3a+6$ | 4 | $\frac{3a+6}{4}$ |

To satisfy $\lambda_1 < \lambda_2 < \lambda_3$, as long as:

$\frac{3a+6}{4}\gt a+1$ is sufficient, which is equivalent to:

$a\lt 2$, so ① is correct.

To let $\lambda_{n+1}\gt \lambda_n$, it is equivalent to:

$S_{n+1}\gt 2S_n$, that is:

$a_{n+1}\gt S_n$

Stronger, just:

$a_n\lt 0(n=1,2,3,...,2024)$ can satisfy $S_n\lt S_{n-1}\lt ...\lt S_1=a\lt a_{n+1}$

Furthermore, we study the necessary and sufficient conditions for ②:

$a_{n+1}=a+2n\gt S_n=na+\frac{n(n-1)}{2}\times 2=na+n(n-1)(n=1,2,..,2024)$

$n=1$, obviously there is: $a_2=a+2\gt a_1=a$

※: $a\lt \frac{-n^2+3n}{n-1}=-n+2+\frac{2}{n-1}(n=2,3,..,2024)$

$u_n=-n+2+\frac{2}{n-1}=1-(n-1-\frac{2}{n-1})(n=2,3,...,2024)$ is a decreasing sequence.

So ② is equivalent to: $a\lt u_{2024}=-2022+\frac{2}{2023}$, ② is correct.

Look again ③:

Based on the analysis of ②, $a\gt 0=u_3$, that is, $n\gt 3$ satisfies $\lambda_{n+1}\lt \lambda_n$, and ③ is correct.

Last look ④:

As long as $a=u_n$ is taken, $\lambda_n=\lambda_{n+1}$ and ④ errors can be satisfied.

In summary, choose ①②③.

## Practice

(End of the first semester of the second grade of Beijing Daxing High School in 2026) (10) It is known that $\{a_n\}$ is an infinite arithmetic sequence in which all terms are not zero, and the tolerance is $d$. Let $c_n=\dfrac{a_1+a_2+\cdots+a_n}{a_n}$, if $a_3\cdot d<0$, then the sequence $\{c_n\}$

(A) There is a maximum term but no minimum term

(B) There is no maximum term and no minimum term

(C) There is a maximum term and a minimum term

(D) There is no maximum term, but there is a minimum term

Let $a_1=a,\ a_n=a+(n-1)d$; then $a_3\cdot d=d(a+2d)<0$.

$c_n=\frac{S_n}{a_n}=\frac{n(a_1+a_n)}{2a_n}=\frac{n}{2}(1+\frac{a_1}{a_n})=\frac{n}{2}(1+\frac{a}{a+(n-1)d})=\frac{n}{2}(1+\frac{1}{1+(n-1)\frac{d}{a}})$

Also $\frac{d}{a}(1+2\frac{d}{a})\lt 0$, so $\frac{d}{a}\in (-\frac{1}{2},0)$.

When n is sufficiently large, $c_n / (\frac{n}{2}) \to 1,c_n\to +\infty$ has no maximum term.

Obviously the sign of $c_n$ is determined by $1+\frac{1}{1+(n-1)\frac{d}{a}}$. If the terms less than 0 in $c_n$ are finite, there must be the smallest $c_n$.

$$
\begin{gathered}
  1+\frac{1}{1+(n-1)\frac{d}{a}}\lt 0\\
\Leftrightarrow1+(n-1)\frac{d}{a}\in (-1,0)\\
\Leftrightarrow (n-1)\in (0,\frac{-2}{\frac{d}{a}})
\end{gathered}$$

Among them, $\frac{-2}{\frac{d}{a}}\gt 4$, such $n$ has only limited possibilities, so $c_n$ has a minimum term.

*PS: Grok didn’t solve this question, but DeepSeek did it, and Musk came out and got beaten)*

## Conclusion

> For the extreme value problem of a sequence, we must not only look at the trend, but also the "turning point". Changes in sign often reveal the existence of extreme values ​​better than monotonic intervals.
> When the trend at infinity of \(c_n\) is established, the extreme value is hidden between the sign interleaving of the finite terms.
> If you can grasp this point, it shows that your understanding of the sequence structure is quite profound.
> As for Grok, it may be good at massive pattern matching, but where mathematical insight is really needed, it's still a bit "smart".
> If you continue to think like this, the final question on the number sequence in the college entrance examination will be nothing more than a paper tiger.