Question 6 • Algebra II
The terms form an arithmetic series, beginning at 19 and ending at 181. Their average is (19 + 181) ÷ 2 = 100, so the sum is 100n. Setting this equal to 10,900 gives n = 10,900 ÷ 100 = 109. Therefore, the series contains 109 numbers, choice C.
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