Exercise 136 Page &nbsp; 329

n= 1

t_1=3( 1)+t_0

MultByOne

a * 1=a

t_1=3+t_0

t_1= 1

1=3+t_0

SubEqn

LHS-3=RHS-3

- 2=t_0

Rearrange equation

t_0= - 2

Now we can complete the equation. t_n=3n+(- 2) ⇔ t_n=3n-2

Recursive

The formula for a recursive rule describing an arithmetic sequence follow a certain format. t(n+1)=t(n)+d, t(0)=a In this formula a is the zeroth term and d is the common difference. When determining the explicit formula, we found that the common difference is 3. We also found that the zeroth term was - 2. With this information we can write the recursive formula. t(n+1)=t(n)+3, t(0)=- 2

b Like in Part A, we will first find the explicit rule and then we will do the recursive.

Explicit

Let's begin by determining what type of sequence this is.

We have a common ratio between consecutive terms, which means this is a geometric sequence. The rule for a geometric sequence follows a certain format. t_n=t_0( b)^nIn the formula, n is the term number, b is the common ratio, and t_0 is the zeroth term. We have already determined that the common ratio is b= 12. t_n=t_0( 1/2)^n To find the zeroth term we can, for example, substitute t_1=3 in the formula and solve for t_0.

t_n=t_0(1/2)^n

n= 1

t_1=t_0(1/2)^1

ExponentOne

a^1=a

t_1=t_0(1/2)

t_1= 3

3=a_0(1/2)

MultEqn

LHS * 2=RHS* 2

6=t_0

Rearrange equation

t_0= 6

Now we can complete the equation. t_n=6(1/2)^n

Recursive

The formula for a recursive rule that describes a geometric sequence follow a certain format. t(n+1)=t(n)* b, t(0)=a In this formula, a is the zeroth term and b is the common ratio. When determining the explicit formula we found that the common difference is 12. We also found that the zeroth term was 6. With this, we can write the recursive formula. t(n+1)=t(n)* 1/2, t(0)=6

c Since this is an arithmetic sequence, we know that a common difference, d, separates consecutive terms.

To get from the first term to the third term, we have to add the common difference twice. With this we can write an equation which will allow us to determine the common difference. 17+2d=2 Let's solve this equation.

17+2d=3

SubEqn

LHS-17=RHS-17

2d=- 14

.LHS /2.=.RHS /2.

d=- 7

Having determined the common difference, we can start writing the explicit formula. t_n= - 7n+t_0 To find the zeroth term we can, for example, substitute t_1=17 in the formula and solve for t_0.

t_n=- 7n+t_0

n= 1

t_1=- 7( 1)+t_0

MultByOne

a * 1=a

t_1=- 7+t_0

t_1= 17

17=- 7+t_0

AddEqn

LHS+7=RHS+7

24=t_0

Rearrange equation

t_0= 24

Now we can complete the equation. t_n=- 7n+24

d Since this is a geometric sequence, we know that a common ratio, b, separates consecutive terms.

To get from the second term to the third term we have to multiply the second term by the common ratio, b. With this we can write an equation with which we can determine the common ratio. 7.2b=8.64 Let's solve this equation.

7.2b=8.64

.LHS /7.2.=.RHS /7.2.

b=1.2

Having determined the common ratio, we can start writing the explicit rule. t_n=t_0( 1.2)^n To find the zeroth term we can, for example, substitute t_2=7.2 in the formula and solve for t_0.

t_n=t_0(1.2)^n

n= 2

t_2=t_0(1.2)^2

t_2= 7.2

7.2=t_0(1.2)^2

CalcPow

Calculate power

7.2=t_0(1.44)

Rearrange equation

t_0(1.44)=7.2

.LHS /1.44.=.RHS /1.44.

t_0=5

Now we can complete the equation. t_n=5(1.2)^n

e Let's illustrate the problem.

To find t(4), we have to determine the explicit rule for this sequence.

To go from the seventh term to the twelfth term we have to add the common difference five times. We can write an equation which will allow us to determine the common difference. 1056+5d=116 Let's solve this equation.

1056+5d=116

SubEqn

LHS-1056=RHS-1056

5d=- 940

.LHS /5.=.RHS /5.

d=- 188

Having determined the common difference, we can start writing the explicit rule. t_n= - 188n+t_0 To find the zeroth term, we can for example, substitute t_7=1056 in the formula and solve for t_0.

t_n=- 188n+t_0

n= 7

t_7=- 188( 7)+t_0

Multiply

t_7=- 1316+t_0