Presentation is loading. Please wait.

Presentation is loading. Please wait.

Lesson 3.12 Applications of Arithmetic Sequences

Similar presentations


Presentation on theme: "Lesson 3.12 Applications of Arithmetic Sequences"β€” Presentation transcript:

1 Lesson 3.12 Applications of Arithmetic Sequences

2 How to identify an arithmetic sequence
In a word problem, look for a constant value being used between each term. This will indicate a common difference, d, and an arithmetic sequence Example: Determine whether each situation has a constant value between each term. The height of a plant grows 2 inches each day. The cost of a video game increases by 10% each month. Johnny receives 5 dollars each week for an allowance.

3 Steps to creating an equation
Create a visual representation of the word problem using the template given. Identify the common difference, 𝑑 and the first term, π‘Ž 1 Determine which formula would best fit the situation (Recursive or Explicit) REMEMBER: Recursive formula helps us get the next term given the previous term while explicit formula gives us a specific term. Plug 𝑑 and π‘Ž 1 in to the formula from step 2. Evaluate the formula for the given term. Interpret the result.

4 Example 1 You visit the Grand Canyon and drop a penny off the edge of a cliff.Β  The distance the penny will fall is 16 feet the first second, 48 feet the next second, 80 feet the third second, and so on in an arithmetic sequence.Β  What is the total distance the object will fall in 6 seconds?

5 Example 1: Sequence Title Picture Description

6 Example 1 You visit the Grand Canyon and drop a penny off the edge of a cliff.Β  The distance the penny will fall is 16 feet the first second, 48 feet the next second, 80 feet the third second, and so on in an arithmetic sequence.Β  What is the total distance the object will fall in 6 seconds? Identify 𝑑 and π‘Ž 1 The given sequence is 16, 48, 80, … d=48βˆ’16=32

7 Example 1 𝑑=32 π‘Žπ‘›π‘‘ π‘Ž 1 =16 Determine which formula would best fit the situation. Since we want the distance after 6 seconds, we will use the explicit formula which is used to find a specific term. Explicit Formula: 𝐴 𝑛 =𝑑 π‘›βˆ’1 + 𝐴 1

8 Example 1 Plug 𝑑 and π‘Ž 1 in to the formula from step 2.
If we plug in 𝑑 and π‘Ž 1 from step 1, we get: 𝐴 𝑛 =32 π‘›βˆ’1 +16 Simplify: 𝐴 𝑛 =32π‘›βˆ’32+16 Distribute 𝐴 𝑛 =32π‘›βˆ’16 Combine Like Terms

9 Example 1 Evaluate the formula for the given value. 𝐴 6 =32(6)βˆ’16
𝐴 𝑛 =32π‘›βˆ’16 Evaluate the formula for the given value. In the problem, we are looking for the total distance after 6 seconds. Therefore, we will plug in 6 to the equation from step 3. 𝐴 6 =32(6)βˆ’16 𝐴 6 =192βˆ’16 𝐴 6 =176

10 𝐴 6 =176 Example 1 Interpret the result
The problem referred to the total distance in feet, therefore: After 6 seconds, the penny will have fallen a total distance of 176 feet.

11 Difference between months
Example 2 Tom just bought a new cactus plant for his office. The cactus is currently 3 inches tall and will grow 2 inches every month. How tall will the cactus be after 14 months? Identify 𝑑 and π‘Ž 1 𝑑=2 π‘Ž 1 =5 Difference between months Height after 1 month

12 Example 2: Sequence Title Picture Description

13 Example 2 𝑑=2 π‘Žπ‘›π‘‘ π‘Ž 1 =5 Determine which formula would best fit the situation. Since we want the distance after 14 months, we will use the explicit formula which is used to find a specific term. Explicit Formula: 𝐴 𝑛 =𝑑 π‘›βˆ’1 + 𝐴 1

14 Example 2 Plug 𝑑 and π‘Ž 1 in to the formula from step 2.
If we plug in 𝑑 and π‘Ž 1 from step 1, we get: 𝐴 𝑛 =2 π‘›βˆ’1 +5 Simplify: 𝐴 𝑛 =2π‘›βˆ’2+5 Distribute 𝐴 𝑛 =2𝑛+3 Combine Like Terms

15 Example 2 Evaluate the formula for the given value. 𝐴 𝑛 =2𝑛+3
𝐴 𝑛 =2𝑛+1 Evaluate the formula for the given value. In the problem, we are looking for the height after 14 months. Therefore, we will plug in 14 to the equation from step 3. 𝐴 𝑛 =2𝑛+3 𝐴 14 = 𝐴 14 =28+3 𝐴 14 =31

16 𝐴 14 =31 Example 2 Interpret the result
The problem referred to the height in inches, therefore: After 14 months, the cactus will be 31 inches tall.

17 Difference between weeks
Example 3 Kayla starts with $25 in her allowance account. Each week that she does her chores, she receives $10 from her parents. Assuming she doesn’t spend any money, how much money will Kayla have saved after 1 year? Identify 𝑑 and π‘Ž 1 𝑑=10 π‘Ž 1 =35 Difference between weeks Amount after 1 week

18 Example 3: Sequence Title Picture Description

19 Example 3 𝑑=10 π‘Žπ‘›π‘‘ π‘Ž 1 =35 Determine which formula would best fit the situation. Since we want the distance after 1 year (Which is ___ weeks), we will use the explicit formula which is used to find a specific term. Explicit Formula: 𝐴 𝑛 =𝑑 π‘›βˆ’1 + 𝐴 1

20 Example 3 Plug 𝑑 and π‘Ž 1 in to the formula from step 2.
If we plug in 𝑑 and π‘Ž 1 from step 1, we get: 𝐴 𝑛 =10 π‘›βˆ’1 +35 Simplify: 𝐴 𝑛 =10π‘›βˆ’10+35 Distribute 𝐴 𝑛 =10𝑛+25 Combine Like Terms

21 Example 3 Evaluate the formula for the given value. 𝐴 𝑛 =10𝑛+25
𝐴 𝑛 = 𝐴 𝑛 =10𝑛+25 Evaluate the formula for the given value. In the problem, we are looking for the amount after 52 weeks. Therefore, we will plug in 52 to the equation from step 3. 𝐴 𝑛 =10𝑛+25 𝐴 52 = 𝐴 52 =520+25 𝐴 52 =545

22 Example 3 Interpret the result 𝐴 52 =545
The problem referred to the amount of money, therefore: After 52 weeks, Kayla will have saved $545.

23 You Try! The sum of the interior angles of a triangle is 180ΒΊ, of a quadrilateral is 360ΒΊ and of a pentagon is 540ΒΊ.Β  Assuming this pattern continues, find the sum of the interior angles of a dodecagon (12 sides).

24 You Try! Sequence Title Picture Description

25 You try! 𝑑=___ π‘Žπ‘›π‘‘ π‘Ž 1 =___ Formula: Plug in 12 Interpret:


Download ppt "Lesson 3.12 Applications of Arithmetic Sequences"

Similar presentations


Ads by Google