Track / Stage 6 / 194 of 400 #1194 of 2000

Problem 1194

National olympiad, first round

Algebra Difficulty 6.3 Prove it

\section*{Problem 1 - 081021}

An arithmetic sequence is a sequence $\left\{a_{1}, a_{2}, \ldots\right\}$ of numbers where all the differences $a_{n+1}-a_{n}(n=1,2, \ldots)$ are equal.

Show that there is exactly one arithmetic sequence such that for every $n=1,2, \ldots$ the sum $S_{n}=a_{1}+a_{2}+\ldots+a_{n}$ of the first $n$ terms is $n^{2}+5 n$ !

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

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If $d$ is the difference of the arithmetic sequence $\left(a_{n}\right)$ , then $a_{n}=a_{1}+(n-1) d$ for $n=1,2, \ldots$ . It follows that

$\begin{aligned} s_{n}= & \left(a_{1}+0 \cdot d\right)+\left(a_{1}+1 \cdot d\right)+\left(a_{1}+2 \cdot d\right)+\ldots+\left(a_{1}+(n-1) \cdot d\right)= \\ & =n \cdot a_{1}+(0+1+2+\ldots+(n-1)) \cdot d=n \cdot a_{1}+\frac{n(n-1)}{2} d \end{aligned}$

It is given that $s_{n}=n^{2}+5 n$ , so $n \cdot a_{1}+\frac{n(n-1)}{2} d=n^{2}+5 n$ for all $n=1,2, \ldots$ . It follows that $a_{1}+\frac{n-1}{2} d=n+5$ for all $n=1,2, \ldots$ .

For $n=1$ , we get $a_{1}=6$ and with $n=2$ then $6+\frac{1}{2} d=7$ , so $d=2$ .

Thus, it is the arithmetic sequence $6,8,10,12, \ldots$ .

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