1980 AMC 12 Problem 11

Attempt Problem 11 of the 1980 AMC 12 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 1980 AMC 12 solutions, or check the answer key.

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11.

If the sum of the first

10

terms and the sum of the first

100

terms of a given arithmetic progression are

100

and

10,

respectively, then the sum of the first

110

terms is

90

-90

110

-110

-100

Write

S_n=\frac n2(2a+(n-1)d)

Use the two known partial sums to find the expression

2a+109d

The two given sums yield

\begin{aligned} 2a+9d&=20,\\ 2a+99d&=\frac15. \end{aligned}

Subtracting gives

90d=-\frac{99}{5},

d=-\frac{11}{50}

and

2a=\frac{1099}{50}.

Therefore

\begin{aligned} S_{110}&=55(2a+109d)\\ &=55(-2)\\ &=-110. \end{aligned}

Therefore, the correct answer is D.

Problem 11 in other years