College Algebra Quiz: Sum Of Arithmetic Series
2 questions · exam conditions
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Sum Of Arithmetic SeriesQuestion 1 of 2

A construction project involves laying concrete blocks in a triangular pattern. The bottom row has 23 blocks, the next row up has 21 blocks, continuing with each row having 2 fewer blocks than the row below it, until reaching a top row with 1 block. What is the total number of blocks used?

144
156
168
180
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College Algebra Quiz

College Algebra Quiz: Sum Of Arithmetic Series

Practice Sum Of Arithmetic Series in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Sum Of Arithmetic Series, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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Question 1

A construction project involves laying concrete blocks in a triangular pattern. The bottom row has 23 blocks, the next row up has 21 blocks, continuing with each row having 2 fewer blocks than the row below it, until reaching a top row with 1 block. What is the total number of blocks used?

  1. 144 (correct answer)
  2. 156
  3. 168
  4. 180
Explanation: This is an arithmetic sequence with a1=23a_1 = 23, d=−2d = -2, and last term an=1a_n = 1. First find nn: 1=23+(n−1)(−2)1 = 23 + (n-1)(-2), so 1=23−2n+2=25−2n1 = 23 - 2n + 2 = 25 - 2n, giving 2n=242n = 24, so n=12n = 12. The sum is S12=122(23+1)=6(24)=144S_{12} = \frac{12}{2}(23 + 1) = 6(24) = 144. Choice B results from miscounting the number of terms (using 13 instead of 12). Choice C comes from using the wrong first term (25 instead of 23). Choice D results from using d=−1d = -1 instead of d=−2d = -2.

Question 2

An arithmetic sequence has the property that the sum of the first 10 terms is 85, and the sum of the next 10 terms (terms 11 through 20) is 285. What is the sum of terms 21 through 30?

  1. 485 (correct answer)
  2. 525
  3. 565
  4. 605
Explanation: Let SnS_n denote the sum of the first nn terms. We have S10=85S_{10} = 85 and S20−S10=285S_{20} - S_{10} = 285, so S20=370S_{20} = 370. For an arithmetic sequence, Sn=n2[2a1+(n−1)d]S_n = \frac{n}{2}[2a_1 + (n-1)d]. From S10=85S_{10} = 85: 102[2a1+9d]=85\frac{10}{2}[2a_1 + 9d] = 85, so 2a1+9d=172a_1 + 9d = 17. From S20=370S_{20} = 370: 202[2a1+19d]=370\frac{20}{2}[2a_1 + 19d] = 370, so 2a1+19d=372a_1 + 19d = 37. Subtracting: 10d=2010d = 20, so d=2d = 2. Then 2a1+18=172a_1 + 18 = 17, so a1=−0.5a_1 = -0.5. The sum of terms 21-30 is S30−S20S_{30} - S_{20}. We have S30=302[2(−0.5)+29(2)]=15[−1+58]=15×57=855S_{30} = \frac{30}{2}[2(-0.5) + 29(2)] = 15[-1 + 58] = 15 \times 57 = 855. Therefore, S30−S20=855−370=485S_{30} - S_{20} = 855 - 370 = 485. Choice B results from using d=2.4d = 2.4 instead of d=2d = 2. Choice C comes from arithmetic errors in the linear system. Choice D results from incorrectly calculating S30S_{30}.