MA.912.AR.10.1

Benchmark Instructional Guide

Connecting Benchmarks/Horizontal Alignment

• MA.912.AR.2.5 
• MA.912.AR.9.10
• MA.912.FL.3

Terms from the K-12 Glossary

• Arithmetic sequence

Vertical Alignment

Previous Benchmarks

• MA.912.AR.2.1
• MA.912.AR.2.2

Next Benchmarks

Purpose and Instructional Strategies

• Instruction focuses on real-world contexts involving arithmetic sequences (MTR.7.1). 
• Students may use a variety of representations to solve problems such as writing out the sequence using the explicit formula. The focus of the benchmark is to understand arithmetic sequences and how they relate to linear equations (MTR.2.1). 
  • In the example given in the benchmark, students are asked to write the total amount of money she has in her account after each month as a sequence. In how many months will she have at least $3,000? The sequence goes as follows: 350, 450, 550, 650, 750, 850, 950, 1050, 1150, 1250, 1350, 1450, 1550, 1650, 1750, 1850, 1950, 2050, 2150, 2250, 2350, 2450, 2550, 2650, 2750, 2850, 2950, 3050, ...
    Once students have written the sequence, they can simply count the number of months it takes to get to at least $3,000. It would take 28 months. 

• Another method to solve this would be to use the explicit formula. In the case of using the explicit formula, students would not need to write as many terms in the sequence since solving the problem this way would not involve counting all the terms until getting to $3,000. When using the explicit formula, arithmetic sequences have a linear relationship where the constant change is the slope while the $y$-intercept is the starting amount. Students can use the equation $y$ = 100$x$ + 250, where $x$ is the number of months and $y$ is the amount of money in her account. To solve this problem students would substitute $3,000 with the $y$-value.
  
  3000 = 100$x$ + 250
  2750 = 100$x$
  27.5 = $x$

• Since she is only putting money in once per month, the $x$ would need to be a whole number. She wouldn’t quite have $3,000 after 27 months so the solution would need to be rounded up to 28 months (MTR.6.1).

• If the $y$-intercept is unknown, students can use point-slope form of a line and any term in the sequence to write the formula for the sequence. For example, in the second month she has $450. Using point-slope form of a line, students get $y$ − 450 = 100($x$ − 2). 
  
  $y$  − 450 = 100($x$ − 2)
  3000 − 450 = 100$x$ − 200
  2550 = 100$x$ − 200
  2750 = 100$x$
  27.5 = $x$

Common Misconceptions or Errors

• Students may have a hard time distinguishing between arithmetic sequences and geometric sequences. 
• Students may have trouble using the equation by confusing the slope and $y$-intercept or not knowing to use point-slope form when the $y$-intercept is unknown. 
• Students may not round correctly when needed depending on the context.

Instructional Tasks

• Albuquerque boasts one of the longest aerial trams in the world. The tram transports people up to Sandia Peak. The table shows the elevation of the tram at various times during a particular ride. The tram increases elevation by 255 feet every minute. After 1 minute, the elevation of the tram is 6,814 feet.
  • Part A. What was the elevation at the beginning of the ride?
  • Part B. If the ride took 15 minutes, what was the elevation of the tram at the end of the ride?

• In a theater, there are 16 seats in the first row and the number of seats in each row after that increase by 2 successively. There are 35 rows in the theater. How many seats does the last row in the theater have?
  a. 65
  b. 78
  c. 84
  d. 86

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