Interpretting Functions: Sequences as Functions and Recursion (CCSS.F-IF.3)
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Common Core: High School - Functions › Interpretting Functions: Sequences as Functions and Recursion (CCSS.F-IF.3)
The arithmetic sequence is 7, 11, 15, 19, ... What is the 5th term?
21
27
23
25
Explanation
This is arithmetic with common difference 4. Using $a_n=a_1+(n-1)d$, we get $a_5=7+4(5-1)=7+16=23$. Choice B (27) uses $7+4\cdot 5$ (off-by-one). Choice A (21) uses a wrong difference of 3. Choice D (25) is the 6th term.
The geometric sequence is 3, 6, 12, 24, ... What is the 5th term?
48
96
24
32
Explanation
This is geometric with ratio $r=2$. Using $a_n=a_1,r^{n-1}$, $a_5=3\cdot 2^{4}=3\cdot 16=48$. Choice B (96) is the 6th term. Choice C (24) is the 4th term. Choice D (32) comes from starting at 1 or using the wrong ratio.
Which recursive definition generates the sequence 2, 5, 8, 11, ...?
$a_0=2$, $a_n=a_{n-1}+3$ for $n\ge 1$
$a_1=2$, $a_n=a_{n-1}+2$ for $n\ge 2$
$a_1=5$, $a_n=a_{n-1}+3$ for $n\ge 2$
$a_1=2$, $a_n=a_{n-1}+3$ for $n\ge 2$
Explanation
The sequence is arithmetic with difference 3 starting at 2, so $a_1=2$ and $a_n=a_{n-1}+3$ for $n\ge 2$. Choice A starts at $n=0$ (indexing shift). Choice B uses the wrong difference. Choice C uses the wrong first term.
A sequence is defined recursively by $a_1=1$, $a_2=2$, and $a_n=a_{n-1}+a_{n-2}$ for $n\ge 3$. What is the 6th term?
8
13
12
11
Explanation
Compute terms: $a_1=1$, $a_2=2$, $a_3=3$, $a_4=5$, $a_5=8$, $a_6=13$. Choice A (8) is the 5th term. Choices C and D come from adding the wrong pair or arithmetic mistakes.
Which recursive definition generates the geometric sequence 3, 9, 27, 81, ...?
$a_1=3$, $a_n=a_{n-1}+3$ for $n\ge 2$
$a_0=3$, $a_n=3a_{n-1}$ for $n\ge 1$
$a_1=3$, $a_n=a_{n-1}^3$ for $n\ge 2$
$a_1=3$, $a_n=3a_{n-1}$ for $n\ge 2$
Explanation
A geometric sequence with ratio 3 and first term 3 satisfies $a_1=3$, $a_n=3a_{n-1}$. Choice A is arithmetic (adds 3). Choice B starts at $n=0$ (indexing shift). Choice C cubes the previous term, which grows too fast.
An arithmetic sequence is defined recursively by $a_1=4$ and $a_n=a_{n-1}+7$ for $n\ge 2$. What is the 5th term, $a_5$?
25
32
39
18
Explanation
Starting at $a_1=4$, add 7 four times to reach $a_5$: $4,11,18,25,32$. So $a_5=32$. Choice A (25) is $a_4$. Choice C (39) adds 7 five times ($4+7\cdot5$). Choice D (18) is $a_3$.
A geometric sequence is defined by $g(0)=3$ and $g(n)=2g(n-1)$ for $n\ge 1$. What is $g(4)$?
24
96
48
12
Explanation
From $n=0$ to $n=4$ there are four doublings: $g(4)=3\cdot 2^4=48$. Choice A (24) is off by one step ($g(3)$). Choice B (96) uses $3\cdot 2^5$. Choice D (12) multiplies by the index instead of repeatedly doubling.
The sequence is 5, 2, $-1$, $-4$, ... Which recursive definition generates this sequence for $n\ge 2$?
$a_1=5$, $a_n=a_{n-1}-3$
$a_0=5$, $a_n=a_{n-1}-3$
$a_1=5$, $a_n=a_{n-1}+3$
$a_1=2$, $a_n=a_{n-1}-3$
Explanation
It is arithmetic with common difference $-3$ and first term 5, so $a_1=5$, $a_n=a_{n-1}-3$. Choice B shifts the start to $n=0$, which would change the indexing. Choice C uses the wrong sign. Choice D starts at 2, not 5.
The sequence $f$ is defined recursively by $f(1)=1$, $f(2)=1$, and $f(n)=f(n-1)+f(n-2)$ for $n\ge 3$. What is $f(6)$?
5
13
6
8
Explanation
Compute terms: $f(1)=1$, $f(2)=1$, $f(3)=2$, $f(4)=3$, $f(5)=5$, $f(6)=8$. Choice A (5) is $f(5)$. Choice B (13) is $f(7)$. Choice C (6) is not a Fibonacci term at this index.
Which recursive definition generates the sequence 81, 27, 9, 3, ...?
$a_1=81$, $a_n=a_{n-1}-3$
$a_1=81$, $a_n=\frac{1}{3}a_{n-1}$
$a_0=81$, $a_n=\frac{1}{3}a_{n-1}$
$a_1=27$, $a_n=\frac{1}{3}a_{n-1}$
Explanation
The sequence divides by 3 each step, so the ratio is $\tfrac{1}{3}$ with first term 81: $a_1=81$, $a_n=\frac{1}{3}a_{n-1}$. Choice A uses subtraction, not multiplication. Choice C starts at $n=0$, shifting indices. Choice D starts at 27, not 81.