1985 AMC 8 Problem 2 — Solution | LIVE by Po-Shen Loh

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1985 AMC 8 Problem 2

Below is the professionally curated solution for Problem 2 of the 1985 AMC 8, 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).

Concepts:arithmetic sequence pairing and grouping

Difficulty rating: 450

2.

What is the value of $90 + 91 + 92 + \cdots + 98 + 99?$

$845$

$945$

$1005$

$1025$

$1045$

Solution:

There are $10$ terms. The average of the first and last is $\dfrac{90 + 99}{2} = 94.5.$

So the sum is $10 \cdot 94.5 = 945.$

Thus, the correct answer is B .

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Problem 2 in Other Years

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