AMC 8 Daily Practice - The Sum of a Finite Arithmetic Series

Complete problem set with solutions and individual problem pages

Problem 7 Medium

37 students line up to count consecutively starting at 1, with each subsequent number increasing by 3. One student wistakenly subtracts 3 instead, resulting in a total sum of 2011. Which student made the counting error?

  • A.

    32

  • B.

    33

  • C.

    34

  • D.

    35

  • E.

    36

Answer:C

If all students counted correctly, the sequence forms an arithmetic progression with: First term a_1 = 1, Common difference d = 3, 37^{\text{th}} term:   a_{37} = a_1 + d(n-1) = 1 + 3 \times (37-1) = 109.

The correct sequence would be 1, 4, 7, 10, \dots, 109. The total sum is: S_{37} = \frac{(a_1 + a_{37}) \times 37}{2} = \frac{(1 + 109) \times 37}{2} = 2035

The discrepancy between the correct sum and actual sum is: 2035 - 2011 = 24

Starting from the erroneous student, each subsequent student's number is 6 less than expected.

The number of affected terms is: \frac{24}{6} = 4

Thus, the error occurs at position: 34

Final result: \boxed{34}

Related Problems
Problem 4 AMC 8 Daily Practice - The Sum of a Finite Arithmetic Series
Problem 5 AMC 8 Daily Practice - The Sum of a Finite Arithmetic Series
Problem 6 AMC 8 Daily Practice - The Sum of a Finite Arithmetic Series
Problem 8 AMC 8 Daily Practice - The Sum of a Finite Arithmetic Series
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