Question

ap precalculus
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date:
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directions: for each of the following, determine if the given sequence is arithmetic, geometric, or neither.

1. ( 12,7,2,-3,-8,cdots )
2. ( 5,10,20,40,cdots )
3. ( 20,10,5,\frac{5}{2},cdots )
4. ( \frac{1}{3},1,\frac{5}{3},\frac{7}{3},3,cdots )
5. ( 1,1,2,3,5,8,13,cdots )
6. ( b_{n}=\frac{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} ).

1. ( a_{3}=7 ) and ( a_{8}=17 )
2. ( a_{2}=-3 ) and ( a_{6}=-9 )
3. ( a_{5}=7 ) and ( d=-4 )
4. ( a_{4}=-1 ) and ( d=\frac{2}{3} )

11.
12.

Explanation:

Step1: Identify the type of sequence for 12,7,2,−3,−8,…

Check the common difference \(d\). \(d = 7 - 12=-5\), \(2 - 7=-5\), \(-3 - 2=-5\), \(-8-(-3)=-5\). Since there is a common difference \(d=-5\), it is an arithmetic sequence.

Step2: Identify the type of sequence for 5,10,20,40,…

Check the common ratio \(r\). \(r=\frac{10}{5} = 2\), \(\frac{20}{10}=2\), \(\frac{40}{20}=2\). Since there is a common ratio \(r = 2\), it is a geometric sequence.

Step3: Identify the type of sequence for 20,10,5,\(\frac{5}{2}\),…

Check the common ratio \(r\). \(r=\frac{10}{20}=\frac{1}{2}\), \(\frac{5}{10}=\frac{1}{2}\), \(\frac{\frac{5}{2}}{5}=\frac{1}{2}\). Since there is a common ratio \(r=\frac{1}{2}\), it is a geometric sequence.

Step4: Identify the type of sequence for \(\frac{1}{3},1,\frac{5}{3},\frac{7}{3},3,\dots\)

Check the common difference \(d\). \(d = 1-\frac{1}{3}=\frac{2}{3}\), \(\frac{5}{3}-1=\frac{2}{3}\), \(\frac{7}{3}-\frac{5}{3}=\frac{2}{3}\), \(3-\frac{7}{3}=\frac{2}{3}\). Since there is a common difference \(d=\frac{2}{3}\), it is an arithmetic sequence.

Step5: Identify the type of sequence for 1,1,2,3,5,8,13,…

Check common difference: \(1 - 1=0\), \(2 - 1 = 1\), \(0
eq1\). Check common ratio: \(\frac{1}{1}=1\), \(\frac{2}{1}=2\), \(1
eq2\). This is the Fibonacci - like sequence (neither arithmetic nor geometric).

Step6: Identify the type of sequence for \(b_{n}=\frac{n + 3}{2}=\frac{1}{2}n+\frac{3}{2}\)

This is in the form of \(a_{n}=a_{1}+(n - 1)d\) (where \(a_{1}=2\), \(d=\frac{1}{2}\)). So it is an arithmetic sequence.

Step7: For \(a_{3}=7\) and \(a_{8}=17\)

Use the formula \(a_{n}=a_{1}+(n - 1)d\). We have \(a_{3}=a_{1}+2d = 7\) and \(a_{8}=a_{1}+7d=17\).
Subtract the first equation from the second: \((a_{1}+7d)-(a_{1}+2d)=17 - 7\), \(5d = 10\), \(d = 2\).
Substitute \(d = 2\) into \(a_{1}+2d = 7\), \(a_{1}+2\times2=7\), \(a_{1}=3\).
So \(a_{n}=3+(n - 1)\times2=2n+1\). Then \(a_{11}=2\times11 + 1=23\).

Step8: For \(a_{2}=-3\) and \(a_{6}=-9\)

Use \(a_{n}=a_{1}+(n - 1)d\). We have \(a_{2}=a_{1}+d=-3\) and \(a_{6}=a_{1}+5d=-9\).
Subtract the first equation from the second: \((a_{1}+5d)-(a_{1}+d)=-9+3\), \(4d=-6\), \(d =-\frac{3}{2}\).
Substitute \(d =-\frac{3}{2}\) into \(a_{1}+d=-3\), \(a_{1}-\frac{3}{2}=-3\), \(a_{1}=-\frac{3}{2}\).
So \(a_{n}=-\frac{3}{2}+(n - 1)\times(-\frac{3}{2})=-\frac{3}{2}n\). Then \(a_{11}=-\frac{3}{2}\times11=-\frac{33}{2}\).

Step9: For \(a_{5}=7\) and \(d=-4\)

Use \(a_{n}=a_{1}+(n - 1)d\). \(a_{5}=a_{1}+4d = 7\), \(a_{1}+4\times(-4)=7\), \(a_{1}=23\).
So \(a_{n}=23+(n - 1)\times(-4)=27-4n\). Then \(a_{11}=27-4\times11=-17\).

Step10: For \(a_{4}=-1\) and \(d=\frac{2}{3}\)

Use \(a_{n}=a_{1}+(n - 1)d\). \(a_{4}=a_{1}+3d=-1\), \(a_{1}+3\times\frac{2}{3}=-1\), \(a_{1}=-3\).
So \(a_{n}=-3+(n - 1)\times\frac{2}{3}=\frac{2}{3}n-\frac{11}{3}\). Then \(a_{11}=\frac{2}{3}\times11-\frac{11}{3}=\frac{11}{3}\).

Answer:

1. Arithmetic
2. Geometric
3. Geometric
4. Arithmetic
5. Neither
6. Arithmetic
7. \(a_{n}=2n + 1\), \(a_{11}=23\)
8. \(a_{n}=-\frac{3}{2}n\), \(a_{11}=-\frac{33}{2}\)
9. \(a_{n}=27-4n\), \(a_{11}=-17\)
10. \(a_{n}=\frac{2}{3}n-\frac{11}{3}\), \(a_{11}=\frac{11}{3}\)