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

Complete problem set with solutions and individual problem pages

Problem 1 Easy

A town hall clock chimes the hour number on every full hour and once on every half-hour. How many chimes will occur in a full 24-hour period?

  • A.

    24

  • B.

    90

  • C.

    156

  • D.

    180

  • E.

    270

Answer:D

Daily chimes:   2 \times \left(1+2+3+\cdots+12+ 12 \right) = 2 \times [ \frac{12 \times \left(1+12\right)}{2} + 12 ] = 180

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Problem 2 Easy

Four campers—Alice, Bob, Charlie, and Diana—play a number-reporting game around a campfire. In Round 1, Alice reports 1, Bob reports 2 and 3, Charlie reports 4, 5, 6, and Diana reports 7, 8, 9, 10. In Round 2, Alice continues with 11 through 15, maintaining the same pattern. How many numbers did Alice report in the first five rounds?

  • A.

    13

  • B.

    17

  • C.

    30

  • D.

    45

  • E.

    50

Answer:D

Alice's reporting pattern follows an arithmetic sequence where the number of terms per round increases by 4 each time: 1, 5, 9, 13, 17 terms for Rounds 1 to 5.

Using the arithmetic series sum formula:

Final result: \boxed{45}

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Problem 3 Easy

A game piece starts at vertex A of a heptagon board labeled A, B, C, D, E, F, G clockwise. Players move the piece 10 times following this rule: on the k-th move, jump k vertices clockwise. Which vertices remain unvisited after 10 moves?

  • A.

    C, E, and F

  • B.

    E, F

  • C.

    C, E

  • D.

    B,C, and F

  • E.

    C, F

Answer:A

Label the vertices as 0 (A), 1 (B), 2 (C), 3 (D), 4 (E), 5 (F), and 6 (G).  

After k moves, the total number of positions traversed is:   S_k = \frac{k(k+1)}{2}  

The final position is determined by computing the remainder Z when S_k is divided by 7:   Z = S_k \bmod 7  

By calculating Z for k=1, 2, \dots, 10, we observe the repeating sequence of remainders:   1, 3, 6, 3, 1, 0, 0, 1, 3, 6.  

Vertices corresponding to remainders 2, 4, and 5 (i.e., C, E, and F) are never visited.  

Final result: \boxed{\text{C, E, and F}}

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Problem 4 Medium

A retirement community has 20 residents with integer ages summing to 1520. The oldest resident is aged >90 but \leq 100, while the remaining 19 residents have consecutive ages. What is the age range of the community?

  • A.

    19

  • B.

    29

  • C.

    30

  • D.

    66

  • E.

    95

Answer:B

Let minimum age be n. The sequence is: \underbrace{n, n+1, \dots, n+18}_{19 \text{ terms}}, \quad m

Sum equation: \frac{19}{2}[2n + 18] + m = 1520 \implies 19n + 171 + m = 1520 \implies m = 1349 - 19n

Given 90 < m \leq 100, solve: 90 < 1349 - 19n \leq 100 \implies 65.7 \leq n < 66.3 \implies n = 66

Thus m = 95, and age range = 95 - 66 = 29.

Final result: \boxed{29}

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Problem 5 Easy

Given the sequence 1, 3, 6, 10, 15, \dots, what is the 20^{\text{th}} term of this sequence?

  • A.

    170

  • B.

    180

  • C.

    190

  • D.

    200

  • E.

    210

Answer:E

This sequence represents triangular numbers with the n-th term formula: T_n = \sum_{k=1}^n k = \frac{n(n+1)}{2}

For n=20: T_{20} = \frac{20 \times 21}{2} = 210

Final result: \boxed{210}

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Problem 6 Easy

An arithmetic sequence has its 3^{\text{rd}} term as 14 and 18^{\text{th}} term as 23. How many of the first 2008 terms are integers?

  • A.

    398

  • B.

    399

  • C.

    400

  • D.

    401

  • E.

    402

Answer:E

Common difference d = \frac{23 - 14}{15} = 0.6.

Integer terms occur when 0.6n \in \mathbb{Z} \implies 0.6 \times 5=3.

In 2008 terms: \left\lfloor \frac{2008-3}{5} \right\rfloor +1 = 402 \text{ integer terms}

Final result: \boxed{402}

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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}

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Problem 8 Easy

What is the value of 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\cdots+\frac{1}{1+2+3+4+\cdots+100}?

  • A.

    \frac{200}{101}

  • B.

    \frac{1}{200}

  • C.

    \frac{101}{200}

  • D.

    \frac{1}{101}

  • E.

    2

Answer:A

1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\cdots+\frac{1}{1+2+3+4+\cdots+100} \\=1+\frac{1}{(1+2) \times 2 \div 2} + \frac{1}{(1+3) \times 3 \div 2} + \cdots +\frac{1}{(1+100) \times 100 \div 2} \\=1+\frac{2}{2 \times 3}+\frac{2}{3 \times 4}+\cdots+\frac{2}{100 \times 101} \\= 1+2 \times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{100}-\frac{1}{101}\right) \\= 1+2 \times\left(\frac{1}{2}-\frac{1}{101}\right) \\= 1+ \frac{99}{101} \\=\frac{200}{101}

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