QUESTION IMAGE
Question
worksheet a. (topic 2.1) arithmetic and geometric sequences
name:
directions: for each of the following, determine if the given sequence is arithmetic, geometric, or neither
- 12, 7, 2, - 3, - 8, ...
- 5, 10, 20, 40, ...
- 20, 10, 5, 5/2, ...
- 1/3, 1, 5/3, 7/3, 3, ...
- 1, 1, 2, 3, 5, 8, 13, ...
- b_n=(n + 3)/2
directions: let a_n be an arithmetic sequence with the following properties. for each of the following, find an expression for a_n, and then find a_11
- a_3 = 7 and a_6 = 17
- a_2=-3 and a_6=-9
- a_3 = 7 and d=-4
- a_4=-1 and d = 2/3
11.
12.
Step1: Recall sequence - type definitions
An arithmetic sequence has a common difference $d=a_{n + 1}-a_{n}$, and a geometric sequence has a common ratio $r=\frac{a_{n+1}}{a_{n}}$.
Step2: Analyze sequence 1 ($12,7,2, - 3,-8,\cdots$)
Calculate the differences: $7 - 12=-5$, $2 - 7=-5$, $-3 - 2=-5$, $-8-(-3)=-5$. Since the common difference $d=-5$, it is an arithmetic sequence.
Step3: Analyze sequence 2 ($5,10,20,40,\cdots$)
Calculate the ratios: $\frac{10}{5}=2$, $\frac{20}{10}=2$, $\frac{40}{20}=2$. Since the common ratio $r = 2$, it is a geometric sequence.
Step4: Analyze sequence 3 ($20,10,5,\frac{5}{2},\cdots$)
Calculate the ratios: $\frac{10}{20}=\frac{1}{2}$, $\frac{5}{10}=\frac{1}{2}$, $\frac{\frac{5}{2}}{5}=\frac{1}{2}$. Since the common ratio $r=\frac{1}{2}$, it is a geometric sequence.
Step5: Analyze sequence 4 ($\frac{1}{3},1,\frac{5}{3},\frac{7}{3},3,\cdots$)
Rewrite the sequence as $\frac{1}{3},\frac{3}{3},\frac{5}{3},\frac{7}{3},\frac{9}{3},\cdots$. Calculate the differences: $\frac{3}{3}-\frac{1}{3}=\frac{2}{3}$, $\frac{5}{3}-\frac{3}{3}=\frac{2}{3}$, $\frac{7}{3}-\frac{5}{3}=\frac{2}{3}$, $\frac{9}{3}-\frac{7}{3}=\frac{2}{3}$. Since the common difference $d = \frac{2}{3}$, it is an arithmetic sequence.
Step6: Analyze sequence 5 ($1,1,2,3,5,8,13,\cdots$)
$1 - 1=0$, $2 - 1 = 1$, differences are not constant. $\frac{1}{1}=1$, $\frac{2}{1}=2$, ratios are not constant. So it is neither arithmetic nor geometric.
Step7: Analyze sequence 6 ($b_{n}=\frac{n + 3}{2}$)
Find $b_{n+1}=\frac{(n + 1)+3}{2}=\frac{n+4}{2}$. Then $b_{n + 1}-b_{n}=\frac{n + 4}{2}-\frac{n + 3}{2}=\frac{(n + 4)-(n + 3)}{2}=\frac{1}{2}$. Since the common difference $d=\frac{1}{2}$, it is an arithmetic sequence.
For the arithmetic - sequence problems:
Problem 7 ($a_{3}=7$ and $a_{6}=17$)
Step1: Use the formula $a_{n}=a_{1}+(n - 1)d$
We know that $a_{3}=a_{1}+2d = 7$ and $a_{6}=a_{1}+5d=17$.
Step2: Solve the system of equations
Subtract the first equation from the second: $(a_{1}+5d)-(a_{1}+2d)=17 - 7$.
$3d=10$, so $d=\frac{10}{3}$.
Substitute $d=\frac{10}{3}$ into $a_{1}+2d = 7$: $a_{1}+2\times\frac{10}{3}=7$, $a_{1}=7-\frac{20}{3}=\frac{21 - 20}{3}=\frac{1}{3}$.
$a_{n}=\frac{1}{3}+(n - 1)\frac{10}{3}=\frac{1+10n - 10}{3}=\frac{10n - 9}{3}$.
$a_{11}=\frac{10\times11-9}{3}=\frac{110 - 9}{3}=\frac{101}{3}$.
Problem 8 ($a_{2}=-3$ and $a_{6}=-9$)
Step1: Use the formula $a_{n}=a_{1}+(n - 1)d$
$a_{2}=a_{1}+d=-3$ and $a_{6}=a_{1}+5d=-9$.
Step2: Solve the system of equations
Subtract the first equation from the second: $(a_{1}+5d)-(a_{1}+d)=-9-(-3)$.
$4d=-6$, so $d =-\frac{3}{2}$.
Substitute $d =-\frac{3}{2}$ into $a_{1}+d=-3$, $a_{1}=-3+\frac{3}{2}=-\frac{3}{2}$.
$a_{n}=-\frac{3}{2}+(n - 1)(-\frac{3}{2})=-\frac{3}{2}-\frac{3}{2}n+\frac{3}{2}=-\frac{3}{2}n$.
$a_{11}=-\frac{3}{2}\times11=-\frac{33}{2}$.
Problem 9 ($a_{3}=7$ and $d=-4$)
Step1: Use the formula $a_{n}=a_{1}+(n - 1)d$
$a_{3}=a_{1}+2d = 7$. Substitute $d=-4$: $a_{1}+2\times(-4)=7$, $a_{1}=7 + 8=15$.
$a_{n}=15+(n - 1)(-4)=15-4n + 4=19-4n$.
$a_{11}=19-4\times11=19 - 44=-25$.
Problem 10 ($a_{4}=-1$ and $d=\frac{2}{3}$)
Step1: Use the formula $a_{n}=a_{1}+(n - 1)d$
$a_{4}=a_{1}+3d=-1$. Substitute $d=\frac{2}{3}$: $a_{1}+3\times\frac{2}{3}=-1$, $a_{1}=-1 - 2=-3$.
$a_{n}=-3+(n - 1)\frac{2}{3}=-3+\frac{2}{3}n-\frac{2}{3}=\frac{2}{3}n-\frac{11}{3}$.
$a_{11}=\frac{2}{3}\times11-\frac{11}{3}=\frac{22 - 11}{3}=\frac{11}{3}$.
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- $a_{n}=\frac{10n - 9}{3}$, $a_{11}=\frac{101}{3}$
- $a_{n}=-\frac{3}{2}n$, $a_{11}=-\frac{33}{2}$
- $a_{n}=19-4n$, $a_{11}=-25$
- $a_{n}=\frac{2}{3}n-\frac{11}{3}$, $a_{11}=\frac{11}{3}$