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AHSME 1980 Problem 11

    If the sum of the first 1010 terms and the sum of the first 100100 terms of a given arithmetic progression are 100100 and 1010, respectively, then the sum of first 110110 terms is: (A) 90(B) −90(C) 110(D) −110(E) −100\text{(A)} \ 90 \text{(B)} \ -90 \text{(C)} \ 110 \text{(D)} \ -110 \text{(E)} \ -100

    Dataset text NuminaMath-1.5. Quoted with source.

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    -110

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