# 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.

Sum of the first 10 terms: $\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 2a+9d=20$ Sum of the first 100 terms: $\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}$

Solving the system, we get $d=-\frac{11}{50}$, $a=\frac{1099}{100}$. The sum of the first 110 terms is $\frac{110}{2}(2a+109d)=55(-2)=-110$

Therefore, $\boxed{D}$.

## See also

 1980 AHSME (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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