Exercise 43 Page 295

Practice makes perfect

a To determine if the sequence is arithmetic or geometric, we have to find out if there is a common difference or a common ratio between consecutive terms.

We have a common difference between consecutive terms, which means this is an arithmetic sequence. The formula for an arithmetic sequence follows a certain format. a_n= mn+a_0 In the formula n is the term number, m is the common difference, and a_0 is the zeroth term. We have already determined that the common difference is m= 12. a_n= 12n+a_0 To find the zeroth term we can, for example, substitute a_1=108 in the formula and solve for a_0.

a_n=12n+a_0

Substitute

n= 1

a_1=12( 1)+a_0

MultByOne

a * 1=a

a_1=12+a_0

Substitute

a_1= 108

108=12+a_0

SubEqn

LHS-12=RHS-12

96=a_0

RearrangeEqn

Rearrange equation

a_0= 96

Now we can complete the equation. a_n=12n+96

b Like in Part A, we will first determine if this is a geometric or an arithmetic sequence.

We have a common ratio between consecutive terms, which means this is a geometric sequence. The formula for a geometric sequence follows a certain format. a_n=a_0( b)^n In the formula, n is the term number, b is the common ratio, and a_0 is the zeroth term. We have already determined that the common ratio is b= 2. a_n=a_0( 2)^n To find the zeroth term we can, for example, substitute a_1= 25 in the formula and solve for a_0.

a_n=a_0(2)^n

Substitute

n= 1

a_1=a_0(2)^1

ExponentOne

a^1=a

a_1=a_0(2)

Substitute

a_1= 2/5

2/5=a_0(2)

DivEqn

.LHS /2.=.RHS /2.

2/5/2=a_0

DivFracD

a/c/b= a/b* c

2/2* 5=a_0

ReduceFrac

a/b=.a /2./.b /2.

1/5=a_0

RearrangeEqn

Rearrange equation

a_0= 1/5

Now we can complete the equation. a_n=1/5(2)^n

c Like in previous parts, let's determine if it is a geometric or an arithmetic sequence.

We have a common difference between consecutive terms which means this is an arithmetic sequence. Let's substitute the common difference, m= - 39, into the general formula for an arithmetic sequence. a_n= - 39n+a_0 To find the zeroth term we can, for example, substitute a_1=3741 in the formula and solve for a_0.

a_n=- 39n+a_0

Substitute

n= 1

a_1=- 39( 1)+a_0

MultByOne

a * 1=a

a_1=- 39+a_0

Substitute

a_1= 3741

3741=- 39+a_0

AddEqn

LHS+39=RHS+39

3780=a_0

RearrangeEqn

Rearrange equation

a_0= 3780

Now we can complete the equation. a_n=- 39n+3780

d Like in previous parts, let's determine if it is a geometric or an arithmetic sequence.

We have a common ratio between consecutive terms which means this is a geometric sequence. Let's substitute the common difference, m= 15, into the general formula for a geometric sequence. a_n=a_0( 1/5)^n To find the zeroth term we can, for example, substitute a_1=117 in the formula and solve for a_0.

a_n=a_0(1/5)^n

Substitute

n= 1

a_1=a_0(1/5)^1

ExponentOne

a^1=a

a_1=a_0(1/5)

Substitute

a_1= 117

117=a_0(1/5)

MultEqn

LHS * 5=RHS* 5