Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
7. Arithmetic Sequences
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Exercise 65 Page 280

Choose an arbitrary value for the common difference and use it to find the value of the fifth term. Use a similar procedure to find the rest of the terms.

Example Function Rule: A(n) = 5 + (n-1)4

Practice makes perfect

We want 25 to be the sixth term of our arithmetic sequence. Recall that the difference between consecutive terms is the common difference d.

Since we do not have any restrictions for the common difference, we can arbitrarily choose any value. Let the common difference be d= 4. Knowing the sixth term and the common difference, we can find the fifth term. To do so, we subtract the common difference from the value of the sixth term. A(5)= 25- 4 ⇔ A(5)= 21 In a similar way, we can find A(4), A(3), A(2), and A(1). A(4)= 21- 4 &⇔ A(4)= 17 A(3)= 17- 4 &⇔ A(3)= 13 A(2)= 13- 4 &⇔ A(2)=9 A(1)= 9- 4 &⇔ A(1)= 5 With this information, we can complete our table.

Now we can use the explicit formula for an arithmetic sequence to write a function rule for our sequence. A(n) = A(1) + (n-1) d In this formula, A(1) is the first term and d is the common difference. The first term of our sequence is 5, and the common difference is 4. We can substitute these values into the above formula. A(n) = 5 + (n-1) 4 This is the function rule we needed. Note that there are infinitely many solutions to this exercise, we illustrated only one example.

Checking Our Answer

Substituting 6 Into the Function Rule
Let's check if the function rule describes a sequence that has 25 as the sixth term. To do so, we will substitute 6 for n into the function rule.

A(n) = 5+ (n-1)4
A( 6)=5 + ( 6-1)4
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Evaluate right-hand side
A(6 )=5 + (5)4
A(6)= 5+20
A(6) = 25

As we can see, the sixth term of the sequence represented by the function rule is 25. Therefore, we are correct.