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Algebra Difficulty 3.2 AMC 10/12 Find the answer

Let SnS_{n} be the sum of the first nn terms of an arithmetic sequence {an}\{a_{n}\}. If the sum of the odd-numbered terms among the first 20172017 terms of {an}\{a_{n}\} is 20182018, then the value of S2017S_{2017} is \_\_\_\_\_\_.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Since SnS_{n} is the sum of the first nn terms of an arithmetic sequence {an}\{a_{n}\}, and the sum of the odd-numbered terms among the first 20172017 terms of {an}\{a_{n}\} is 20182018,
we have Sodd=a1+a3+a5++a2017=1009×(a1+a2017)×12=2018S_{odd} = a_{1} + a_{3} + a_{5} + \ldots + a_{2017} = 1009 \times (a_{1} + a_{2017}) \times \frac{1}{2} = 2018, which gives a1+a2017=4a_{1} + a_{2017} = 4.
Then, S2017=20172(a1+a2017)=2017×2=4034S_{2017} = \frac{2017}{2}(a_{1} + a_{2017}) = 2017 \times 2 = 4034.
Therefore, the answer is 4034\boxed{4034}.
This problem examines the formula for the sum of an arithmetic sequence Sn=n2(a1+an)S_{n} = \frac{n}{2}(a_{1} + a_{n}). First, by using Sodd=a1+a3+a5++a2017=1009×(a1+a2017)×12=2018S_{odd} = a_{1} + a_{3} + a_{5} + \ldots + a_{2017} = 1009 \times (a_{1} + a_{2017}) \times \frac{1}{2} = 2018, we find that a1+a2017=4a_{1} + a_{2017} = 4. Then, calculating S2017=20172(a1+a2017)=2017×2=4034S_{2017} = \frac{2017}{2}(a_{1} + a_{2017}) = 2017 \times 2 = 4034 is straightforward.
This question tests the formula for the sum of an arithmetic sequence and calculation skills, and is considered a medium-level problem.

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