What's the recursive rule for this sequence? 75, 100, 125, 156,

6 answers
Question:

What's the recursive rule for this sequence? 75, 100, 125, 156,

Answers

1. Let [tex]s_n[/tex] be the number of seats in the [tex]n[/tex]-th row. The number seats in the [tex]n[/tex]-th row relative to the number of seats in the [tex](n-1)[/tex]-th row is given by the recursive rule

[tex]s_n=s_{n-1}+5[/tex]

Since [tex]s_1=21[/tex], we have

[tex]s_2=s_1+5[/tex]
[tex]s_3=s_2+5=s_1+2\cdot5[/tex]
[tex]s_4=s_3+5=s_1+3\cdot5[/tex]
[tex]\cdots[/tex]
[tex]s_n=s_{n-1}+5=\cdots=s_1+(n-1)\cdot5[/tex]

So the explicit rule for the sequence [tex]s_n[/tex] is

[tex]s_n=21+5(n-1)\implies s_n=5n+16[/tex]

In the 15th row, the number of seats is

[tex]s_{15}=5(15)+16=91[/tex]

2. Let [tex]p_n[/tex] be the amount of profit in the [tex]n[/tex]-th year. If the profits increase by 6% each year, we would have

[tex]p_2=p_1+0.06p_1=1.06p_1[/tex]
[tex]p_3=1.06p_2=1.06^2p_1[/tex]
[tex]p_4=1.06p_3=1.06^3p_1[/tex]
[tex]\cdots[/tex]
[tex]p_n=1.06p_{n-1}=\cdots=1.06^{n-1}p_1[/tex]

with [tex]p_1=40,000[/tex].

The second part of the question is somewhat vague - are we supposed to find the profits in the 20th year alone? the total profits in the first 20 years? I'll assume the first case, in which we would have a profit of

[tex]p_{20}=1.06^{19}\cdot40,000\approx121,024[/tex]

3. Now let [tex]p_n[/tex] denote the number of pushups done in the [tex]n[/tex]-th week. Since [tex]3\cdot4=12[/tex], [tex]12\cdot4=48[/tex], and [tex]48\cdot4=192[/tex], it looks like we can expect the number of pushups to quadruple per week. So,

[tex]p_n=4p_{n-1}[/tex]

starting with [tex]p_1=3[/tex].

We can apply the same reason as in (2) to find the explicit rule for the sequence, which you'd find to be

[tex]p_n=4^{n-1}p_1\implies p_n=4^{n-1}\cdot3[/tex]

A

this is a geometric sequence since there exists a common ratio r between the terms

r = [tex]\frac{21}{7}[/tex] = [tex]\frac{63}{21}[/tex] = [tex]\frac{189}{63}[/tex] = 3

B

to obtain the next term in the sequence multiply the previous term by 3

[tex]a_{n+1}[/tex] = 3 [tex]a_{n}[/tex] ← recursive rule

C

the n th term of a geometric sequence is

[tex]a_{n}[/tex] = [tex]a_{1}[/tex][tex]r^{n-1}[/tex]

where [tex]a_{1}[/tex] is the first term in the sequence

[tex]a_{n}[/tex] = 7 × [tex]3^{n-1}[/tex] ← explicit rule

Answer 1: The correct option is C ). A(n)=A(n-1)+4.3 , where A(1)=(−22.7)

Answer 2: The correct option is A ). A(n)=18−3.5n

Answer 3: The correct option is D ). A(1)=4; A(n)=A(n−1)+3

Answer 4: The correct option is B ). A(n)=17n-1

Step-by-step explanation:

An arithmetic sequence is given by a,a+d,a+2da+(n-1)d

a is first term of sequence and d is interval of sequence

Where A(n)=a+(n-1)d

The recursive rule for the sequence is given by

A(n)=A(n-1)+d

The explicit rule for the sequence is given by

A(n)=a+(n-1)d

Answer 1:

Given sequence is −22.7, −18.4, −14.1, −9.8, −5.5

One comparing the sequence

we get, a=(-22.7) and a+d=(-18.4)

Interval of sequence d=(a+d)-(a)

d=(-18.4)-(-22.7)

d=4.3

The recursive rule for the sequence is given by

A(n)=A(n-1)+d

A(n)=A(n-1)+4.3

The correct option is C ). A(n)=A(n-1)+4.3 , where A(1)=(−22.7)

Note : The correct option is D is correct, if we write recursive rule as

A(n)=A(n+1)-d

A(n)=A(n+1)-4.3

Answer 2:

Given sequence is 14.5, 11, 7.5, 4, 0.5

One comparing the sequence

we get, a=14.5 and a+d=11

Interval of sequence d=(a+d)-(a)

d=(11)-(14.5)

d=(-3.5)

The explicit rule for the sequence is given by

A(n)=a+(n-1)d

A(n)=14.5+(n-1)(-3.5)

A(n)=14.5-3.5n+3.5

A(n)=18-3.5n

The correct option is A ). A(n)=18−3.5n

Answer 3:

Given explicit rule for a sequence is A(n)=3n+1

For A(1):

A(n)=3n+1

A(1)=3(1)+1=4

For A(n-1):

a(n)=3n+1

A(n-1)=3(n-1)+1

A(n-1)=3n-3+1

A(n-1)=3n-2

The recursive rule for the sequence is given by

A(n)=A(n-1)+d

3n+1=3n-2+d

d=3

The recursive rule for the sequence is A(n)=A(n-1)+3

The correct option is D ). A(1)=4; A(n)=A(n−1)+3

Answer 4:

A supermarket finds that the number of boxes of a new cereal sold increases each week.

In the first week, 16 boxes of the cereal were sold

In the second week, 33 boxes of the cereal were sold

In the third week, 50 boxes of the cereal were sold

An arithmetic sequence is given by a,a+d,a+2da+(n-1)d

Here,

a=16 and a+d=33

Hence d=(a+d)-(a)=33-16=17

The explicit rule for the sequence is given by

A(n)=a+(n-1)d

A(n)=16+(n-1)17

A(n)=16+17n-17

A(n)=17n-1

The correct option is B ). A(n)=17n-1

ANSWER

See explanation

EXPLANATION

Question 1:

The third term of the arithmetic sequence is :

14=a+2d...(1)

The twelveth term is

59=a+11d...(2)

Subtract equation (1) from (2)

45=9d

This implies that

d=5

a=14-2(5)=4

The explicit rule is;

[tex]a_{n}=4 + 5(n - 1)[/tex]

[tex]a_{n}=4 + 5n -5[/tex]

[tex]a_{n} = 5n -1[/tex]

Recursive formula:

[tex]a_{n}=a_{n - 1} + 5[/tex]

Question 2

The geometric sequence has the fourth term to be 2 and the common ratio to be r=⅓

This implies that,

[tex]a {( \frac{1}{3} })^{3} = 2[/tex]

This implies that,

[tex]\frac{a}{27} = 2[/tex]

[tex]a = 54[/tex]

The explicit rule:

[tex]a_n=54 {( \frac{1}{3} })^{n - 1}[/tex]

The recursive rule is

[tex]a_n=( \frac{1}{3} )a_{n-1}[/tex]

where,

[tex]a_1 = 54[/tex]

I'm having trouble with this type of math to so your not alone

Step-by-step explanation:

The recursive rule for the sequence is a[tex]_{1}[/tex] = 75; a[tex]_{n}[/tex] = a[tex]_{n-1}[/tex] + 25

Step-by-step explanation:

the recursive rule of an arithmetic function is a[tex]_{1}[/tex] = first term; a[tex]_{n}[/tex]= a[tex]_{n-1}[/tex] + d, where

a[tex]_{1}[/tex] is the first terma[tex]_{n-1}[/tex] is the nth termd is the common difference between each 2 consecutive terms

∵ The given is 75, 100, 125, 150, .......

∵ 100 - 75 = 25

∵ 125 - 100 = 25

∵ 150 - 125 = 25

∴ The common difference is 25

∴ d = 25

∵ The first term is 75

∴ a[tex]_{1}[/tex] = 75

→ Substutute the values of a1 and d in the formula above

∴ The recursive rule for the sequence is a[tex]_{1}[/tex] = 75; a[tex]_{n}[/tex] = a[tex]_{n-1}[/tex] + 25

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