✏️ARITHMETIC SERIES
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Question: In a display of cans, there are 18 cans in the bottom row, 17 in the row above, 16 in th next row, and so on. How many cans are there in all if there are 6 rows?
Answer: [tex] \Large \underline{ \rm \green{ \: D. \: 93 \: }}[/tex]
Solution: Each rows will form a sequence in which -1 is added to the next rows, that would be our common difference.
- [tex]d = a_2 - a_1 = 17 - 18 = \text - 1[/tex]
- [tex]d = a_3 - a_2 = 16 - 17 = \text -1[/tex]
- The first term of the sequence is 18 since that is the first row. Find the sum of the first 6 terms of the sequence or the total number of cans in all.
[tex] \begin{aligned} & \bold{ \color{lightblue}Formula:} \\ & \boxed{S_n = \frac{n}{2} \big[2a_1 + d(n - 1) \big]} \end{aligned}[/tex]
- [tex] \begin{aligned}{S_6 = \frac{6}{2} \big[2(18) + ( \text - 1)(6 - 1) \big]} \end{aligned}[/tex]
- [tex] \begin{aligned}{S_6 = \frac{6}{2} \big[2(18) - 1(6 - 1) \big]} \end{aligned}[/tex]
- [tex] \begin{aligned}{S_6 = \frac{6}{2} \big[2(18) - 1(5) \big]} \end{aligned}[/tex]
- [tex] \begin{aligned}{S_6 = \frac{6}{2} \big[36 - 5 \big]} \end{aligned}[/tex]
- [tex] \begin{aligned}{S_6 = 3 \big[31\big]} \end{aligned}[/tex]
- [tex] \begin{aligned}{S_6 = 93} \end{aligned}[/tex]
- Therefore, there are 93 cans in all.
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