Sum of Arithmetic Series - College Algebra

1. 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?
   Answer: 485 Let $S_n$ denote the sum of the first $n$ terms. We have $S_{10} = 85$ and $S_{20} - S_{10} = 285$, so $S_{20} = 370$. For an arithmetic sequence, $S_n = \frac{n}{2}[2a_1 + (n-1)d]$. From $S_{10} = 85$: $\frac{10}{2}[2a_1 + 9d] = 85$, so $2a_1 + 9d = 17$. From $S_{20} = 370$: $\frac{20}{2}[2a_1 + 19d] = 370$, so $2a_1 + 19d = 37$. Subtracting: $10d = 20$, so $d = 2$. Then $2a_1 + 18 = 17$, so $a_1 = -0.5$. The sum of terms 21-30 is $S_{30} - S_{20}$. We have $S_{30} = \frac{30}{2}[2(-0.5) + 29(2)] = 15[-1 + 58] = 15 \times 57 = 855$. Therefore, $S_{30} - S_{20} = 855 - 370 = 485$. Choice B results from using $d = 2.4$ instead of $d = 2$. Choice C comes from arithmetic errors in the linear system. Choice D results from incorrectly calculating $S_{30}$.
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?
   Answer: 144 This is an arithmetic sequence with $a_1 = 23$, $d = -2$, and last term $a_n = 1$. First find $n$: $1 = 23 + (n-1)(-2)$, so $1 = 23 - 2n + 2 = 25 - 2n$, giving $2n = 24$, so $n = 12$. The sum is $S_{12} = \frac{12}{2}(23 + 1) = 6(24) = 144$. 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 $d = -1$ instead of $d = -2$.

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