Answer:
Recursive:
[tex]f(1)=35, f(n)=f(n-1)+10[/tex]
Explicit:
[tex]f(n)=35+10(n-1)[/tex]
And the 20th term is 225.
Step-by-step explanation:
We have the sequence:
35, 45, 55, 65.
Notice that each subsequent term is 10 more than the previous term.
Therefore, our common difference is (+)10.
Recursive Rule:
The standard format for the recursive rule is:
[tex]f(n)=a, f(n)=f(n-1)+d[/tex]
Where a is the initial term and d is the common difference.
From our sequence, we know that a the initial term is 35.
And as determined, our common difference d is 10.
Substitute. Hence, our recursive rule is:
[tex]f(1)=35, f(n)=f(n-1)+10[/tex]
Explicit Rule:
The standard format for the explicit rule is:
[tex]f(n)=a+d(n-1)[/tex]
Where a is the initial term and d is the common difference. So, let’s substitute 35 for a and 10 for d. Hence, our explicit formula is:
[tex]f(n)=35+10(n-1)[/tex]
Now, let’s find the 20th term. We will utilize the explicit rule since the recursive rule can get tedious. Substitute 20 for n because we would like to 20th term. Thus:
[tex]f(20)=35+10(20-1)[/tex]
Evaluate:
[tex]\begin{aligned} f(20)&=35+10(19) \\ f(20)&=35+190 \\ f(20)&=225 \end{aligned}[/tex]
Hence, the 20th term is 225.
