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Austria competition problems, 2010

Algebra Difficulty 6.7 National Olympiad Prove it Austria

A sequence $\langle a_n \rangle$ with $a_n = a + nd$ is called an arithmetic sequence. The sequence $\langle b_n \rangle$ with $b_n = \sum_{k=0}^{n} a_k$ is called an arithmetic sequence of second degree. Let $a$ and $d$ be positive integers.

We consider all such arithmetic sequences of second degree containing the number $2010$ . What is the highest possible index $n$ if $b_n = 2010$ ? Determine all possible arithmetic sequences $\langle a_n \rangle$ , for which $b_n = 2010$ holds for this index.

Solution

Since $a_k = a + k d$ , we have $b_n = (n+1) \cdot a + \frac{n(n+1)}{2} \cdot d$ . If we assume $b_n = 2010$ , we therefore obtain
$a = \frac{4020 - n(n + 1)d}{2(n + 1)} = \frac{2010}{n + 1} - \frac{dn}{2}.$
Since both $a$ and $d$ are positive, we see from the first fraction, that $n(n + 1) < 4020$ must hold. We therefore have $n(n + 1) < 4020 < 4225 = 65^2$ , and thus $n < 65$ . Furthermore, $n + 1$ must divide $4020$ . Since $4020 = 3 \cdot 4 \cdot 5 \cdot 67$ holds, the largest divisor of $4020$ less than $65$ is $60$ . The largest possible value for $n + 1$ is therefore $60$ , and we have $n = 59$ .

Substituting this value yields $a = \frac{67-59d}{2}$ . We see that $d$ must be odd and less than $2$ , and we therefore have the unique solution $d = 1$ , which yields $a = 4$ .

The only possible sequence $\langle a_n \rangle$ , for which $b_{59} = 2010$ is therefore the sequence $\langle 4, 5, 6, \dots \rangle$ .

qed

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