1980 AHSME Problems/Problem 11

Problem

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 first $110$ terms is:

$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$

Solution

Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.

The sum of the first $10$ terms is $\frac{10}{2}(2a+9d)=100$ . This equation can be simplified to $2a+9d=20$ .

The sum of the first $100$ terms is $\frac{100}{2}(2a+99d)=10$ . This equation can be simplified to $2a+99d=\frac{1}{5}$ .

Solving the system of these two equations yields $(a, d) = \left(\frac{1099}{100}, -\frac{11}{50}\right)$ . Therefore, the sum of the first $110$ terms is $\frac{110}{2}(2a+109d)=55\left(2\cdot\frac{1099}{100} + 109 \cdot -\frac{11}{50}\right) = 55 \cdot -2 = \boxed{(\textbf{D}) -110}$ .

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1980 AHSME Problems/Problem 11

Problem

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