Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
6. Arithemetic Sequences
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Exercise 55 Page 206

Use the formula for the nth term of an arithmetic sequence and the given terms to solve for d.

f(n)=6n + 17

Practice makes perfect

We can define an arithmetic sequence as a function of n. f(n)=a_1+(n-1)d This function generates the terms of any arithmetic sequence given a certain starting value a_1 and common difference d. The variable n is the independent variable and will always be an integer. It represents the position of the term within the sequence. Therefore, the arithmetic sequence will have the following terms. f(1), f(2), f(3), ...From the given diagram, we can identify a few elements in our arithmetic sequence. f(1)&=23 ⇒ a_1=23 f(4)&=41 ⇒ a_4=41 f(12)&=89 ⇒ a_(12)=89 We already know that the first term a_1 is 23. f(n)= 23+(n-1)d Now, to find the value of d, we will substitute the information for the 4th term into the equation. In other words, we will substitute 4 for n and 41 for f(n).

f(n)=23+(n-1)d
41=23+( 4-1)d
â–¼
Solve for d
41=23+3d
18=3d
6=d
d=6

The common difference is 6. Let's use this information to write our function. f(n)=23+(n-1) 6 Finally, let's distribute 6 in the equation to simplify the right-hand side.

f(n)=23+(n-1)6
f(n)=23+6n-6
f(n)=6n+17