1985 AMC 8 Problem 2

Attempt Problem 2 of the 1985 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1985 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

2.

What is the value of 90+91+92+⋯+98+99?90 + 91 + 92 + \cdots + 98 + 99?
845845
945945
10051005
10251025
10451045
Answer: B
Concepts:arithmetic sequencepairing and grouping
Small hint:
There are 1010 terms, from 9090 up to 9999
Big hint:
The sum equals the average of the first and last term times the number of terms
Solution:
There are 1010 terms. The average of the first and last is 90+992=94.5.\dfrac{90 + 99}{2} = 94.5.
So the sum is 10⋅94.5=945.10 \cdot 94.5 = 945.
Thus, the correct answer is B .
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