INTERNET MAT 117

SOLUTION for the REVIEW

(1) Let us consider the circle with equation

x2 + y2 + 2x + 3y +3

4= 0.

(a) Find the standard form of the equation of the circle given above.(i) Group the x and y terms together and take the constant term

to the other side.

x2 + 2x + y2 + 3y = −3

4

(ii) Complete the square by adding the square of half of the coef-ficients of x and y to both sides of the equation. Add 1 and32

22 = 94

to both sides of the equation.

x2 + 2x + 1 + y2 + 3y +9

4= 1 +

9

4− 3

4

(iii) Complete the square, combine the constants. See the stan-dard form below:

(x + 1)2 + (y +3

2)2

=10

4=

5

2

(b) Find the center and the radius of the circle given above.

Center: (−1,−3

2).

Radius:√

5/2 = 1.58113883.

(2) (a) Find the slope intercept form of the line passing through (−2, 5)and parallel to the line 5x− 3y = 8.

(i) Find the slope intercept form of the line 5x − 3y = 8 byexpressing y in terms of x.

−3y = −5x + 8

y =5

3x− 8

3

Thus, the slope of the line 5x− 3y = 8 is5

3. Hence, the slope

of a parallel line to the line 5x− 3y = 8 is also5

3.

2

(ii) Now we have to find the equation of the line passing through

the point (−2, 5) that has the slope5

3. The slope intercept

form of any line with slope5

3can be written in the form of

y =5

3x + b.

The line passing through the point (−2, 5). Substitute x = −2and y = 5 and solve the equation for b.

5 = −10

3+ b.

b =25

3So the equation of the line passing through the point (−2, 5)and has the slope 5

3is

y =5

3x +

25

3.

(b) Find the slope intercept form of the line passing through (−2, 5)and perpendicular to the line 5x− 3y = 8.

(i) Find the slope intercept form of the line 5x − 3y = 8 byexpressing y in terms of x

−3y = −5x + 8

y =5

3x− 8

3

Thus, the slope of the line 5x− 3y = 8 is5

3. Hence, the slope

of a perpendicular line to the line 5x− 3y = 8 is −3

5.

(ii) Now we have to find the equation of the line passing through

the point (−2, 5) that has the slope −3

5. The slope intercept

form of any line with slope −3

5can be written in the form of

y = −3

5x + b.

The line passing through the point (−2, 5). Substitute x = −2and y = 5 and solve the equation for b.

5 =6

5+ b.

b =19

5

3

So, the equation of the line passing through the point (−2, 5)

and that the slope −3

5is

y = −3

5x +

19

5.

(3) Find the domain of the following functions. Give your answer in intervalnotation.

(a) g(x) =

√x− 1

5− x(i) The denominator cannot be zero , so x 6= 5.(ii) The expression under the radical cannot be negative, so 1 ≤ x.

Hence 1 ≤ x and x 6= 5. With interval notation,

[1, 5) ∪ (5,∞).

(b) f(x) =x + 1

4− x2

(i) The denominator cannot be zero , so x2− 4 6= 0 that is x 6= 2and x 6= −2. With interval notation

(∞,−2) ∪ (−2, 2) ∪ (2,∞).

(4) A small business buys a computer for $ 4,000. After 4 years the valueof the computer is expected to be $ 200. For accounting purposes, thebusiness uses linear depreciation to assess the value of the computer ata given time. This means the if V is the value of the computer at thetime t, then a linear equation is used to relate V and t.

(a) Find a linear equation that relates V and t.

Slope of the line passing through the points (0, 4000) and (4,200)is −950. The equation of the line passing through the give pointswith slope −950 is

V (t) = −950t + 4000

(b) Find the depreciated values of the computer 3 years from the dateof purchase.

Substitute t = 3 years in to the equation obtained in part a. ThenV = 1150. So, it is $1,150.

(5) Let f(x) = x2 − 1 and g(x) =√

x + 2

(a) Find (f

g)(x) and give the domain of

f

g.

(f

g)(x) =

x2 − 1√x + 2

.

4

(i) The denominator can not be zero and the expression can notbe negative under the radical, so x > −2.

(ii) The domain of (f

g)(x) is (−2,∞) since x > −2.

(b) Find (f ◦ g)(x) and (g ◦ f)(x)

(f ◦ g)(x) = x + 2− 1 = x + 1(g ◦ f)(x) =

√x2 − 1 + 2 =

√x2 + 1

(6) Let f(x) = −2x + 4.

(a) Find f−1(x).

y = −2x + 4(i) Solve the equation for x. (Express x in terms of y)

y − 4 = −2xy − 4

−2= x or

4− y

2= x

(ii) Interchange the variables x and y.

4− x

2= y

f−1(x) =4− x

2(b) Graph f and f−1 on the same coordinate axis.

−6 −4 −2 0 2 4 6−6

−4

−2

0

2

4

6

y = −2x+4y = (4−x)/2

5

(7) Use the graph of f(x) below to sketch the graph of g(x) it if is knownthat g(x) = f(x − 1) + 20. Draw g(x) on the same set of axes as thegiven graph.

−4 −3 −2 −1 0 1 2 3 4 5 6 7−40

−20

0

20

40

60

80

f(x) f(x−1)

f(x−1)+20

6

(8) Consider the graph of the function f(x) below. Answer the followingquestions:

−4 −3 −2 −1 0 1 2 3 4 5 6−3

−2

−1

0

1

2

3

4

5

(a) Use interval notation to state the domain of f(x).

[−2, 4]

(b) Use interval notation to state the range of f(x).

[−1, 4]

(c) State the interval where the function f(x) increases.

[0, 1)

(d) State the interval where the function f(x) decreases.

[−2, 0] ∪ [1, 4]

7

(e) State the values of f(1) and f(4).

f(1) = 2 f(4) = −1

(9) Let P (x) = x3 − 3x2 − 9x− 5

(a) Determine all the possible rational zeros for P (x).

By the Rational Zeros Theorem if pq

is a zero of P (x) then p divides

5 and q divides 1, so pq

in the form of

factor of 5

factor of 1.

The factors of 5 are 1,-1, 5, -5 and the factors of 1 are 1 and -1.Thus, the possible rational zeros p

qare 1, -1, 5, -5.

(b) Factor the polynomial P (x) completely using the procedure of longdivision or synthetic division.

5 is a zero, that is P(5)=0. Using synthetic division ( or long divi-sion) we obtainx3 − 3x2 − 9x− 5 = (x− 5)(x2 + 2x + 1) = (x− 5)(x + 1)2.

(c) Find all the zeros of P (x) and state the multiplicity for each zero.Give the exact answer. No decimals.

x = 5 multiplicity 1.x = −1 multiplicity 2.

(d) Find where P (x) approaches when x →∞ and whenx → −∞.

P (x) →∞ as x →∞.P (x) → −∞ as x → −∞.

(e) Sketch the graph of P (x) indicating the zeros.

8

−6 −4 −2 0 2 4 6−35

−30

−25

−20

−15

−10

−5

0

5

10

15

20

x = 5x = −1

(10) Consider the polynomial P (x) = x(x + 2)2(x− 3)3.

(a) Find all zeros of P (x) and state their multiplicities.

x = −2 multiplicity 2.x = 0 multiplicity 1.x = 3 multiplicity 3.

(b) Where P (x) approaches when x →∞ and when x → −∞.

P (x) →∞ as x →∞.P (x) →∞ as x → −∞.

(c) Sketch the graph of the function P (x) indicating the x-intercepts(the zeros).

9

−4 −3 −2 −1 0 1 2 3 4 5 6 7−100

−80

−60

−40

−20

0

20

40

60

80

100

x=−2 x=0 x=3

(11) Evaluate the expression1− 3i

2 + 5iand write the result in the a + bi form .

Multiply both the numerator and denominator by the conjugate ofthe denominator which is 2−5i. (Note that we did not change the value

of the fraction because2− 5i

2− 5i= 1

1− 3i

2 + 5i=

(1− 3i)

(2 + 5i)· (2− 5i)

(2− 5i)=−13− 11i

29= −13

29− 11

29i

So, a = −13

29and b = −11

29.

(12) Find all the real zeros of the polynomial x3+2x2−10x−15. Use the qua-dratic formula if it is necessary. Use radicals, don’t use decimals.(Hint:first find the rational zeros.)

Possible rational zeros: 1,-1, 3, -3, 5, -5, 15, -15.x = 3 is a zero, P (3) = 0. Using synthetic divisionx3 + 2x2 − 10x− 15 = (x− 3)(x2 + 5x + 5).Using the quadratic formula for x2 +5x+5 = 0 with a = 1, b = 5 and

c = 5 we get the zeros are x = 5, x =−5 +

√5

2and x =

−5−√5

2.

10

(13) Find all the complex zeros of x2+2x+2. Use radicals don’t use decimals.

Using the quadratic formula with a = 1, b = 2 and c = 2 we obtain

x =−2 +

√−4

2= −1 + i and x =

−2−√−4

2= −1− i.

(14) Find the quotient (Q(x)) and the remainder (R(x)) using the long divi-sion.

3x4 − 5x3 + 4x + 3

x2 + x + 3.

+3x2 −8x −1

x2 +x +3 3x4 −5x3 +0x2 +4x +33x4 +3x3 +9x2

−8x3 −9x2 +4x +3−8x3 −8x2 −24x

−x2 +28x +3−x2 −x −3

29x +6

Q(x) = 3x2 − 8x− 1.R(x) = 29x + 6.

(15) An object is projected upward from the top of a building. The height ofthe object in meters is described by the function h(t) = −4.9t2 + 40t +10, where t is in seconds and corresponds to the moment the object isprojected.

(a) Determine the height of the building.

In the t = 0 moment the object is in the top of the building. So,when t = 0 then h = 10 meters. Thus, the height of the building is10 meters.

(b) Algebraically determine for what value of t the object reaches themaximum height and determine this maximum height.(Hint use thevertex formula.) Check your answer with your calculator.

Note that the graph of the height describes by the moving objectis an open downward parabola. So, the maximum height is at thevertex of the parabola. Using the vertex formula with b = 40 anda = −4.9 the object reaches the maximum height when t = − b

2a=

4.081632653 seconds. Substituting t = 4.081632653 seconds intoh(t) = −4.9t2 + 40t + 10. we obtain that the maximum height is91.63265306 meters.

11

(c) Algebraically determine when the object reaches the ground. (Hintuse the quadratic formula). Check your answer with your calculator.

The object reaches the ground when h = 0. So, we have to substi-tute h = 0 and solve the equation for t using the quadratic formulawith c = 10, b = 40 and a = −4.9. Since the variable t describesthe time we just consider the positive solution. The solution for theequation 0 = −4.9t2 + 40t + 10 is t = 8.406044918 seconds.

(16) The function p(x) = −x2+46x−360 models the daily profit in hundredsof dollars for a company that manufactures x computers daily. (You mayshow your work algebraically or graphically, which includes a sketch ofthe graph.)

(a) How many computers should be manufactured each day to maxi-mize profit?

Note that the graph of the function which describes the daily profitis an open downward parabola. So, the maximum profit is at thevertex of the parabola. Using the vertex formula with b = 46 anda = −1 the manufacturer reaches the maximum profit when x =− b

2a= 46 . So, x = 23 computers should be manufactured each day

to maximize profit.

(b) What is the maximum daily profit?

Substituting x = 46 into p(x) = −x2 + 46x − 360. we obtain thatthe maximum profit is $ 16,900, since the profit function describesthe profit in hundreds of dollars.

(17) (a) Consider the rational function R(x) =x2 − x− 6

2x2 − 2.

(i) Find the vertical asymptotes of R(x) if there is any.

Factor the numerator and the denominator if you can.

x2 − x− 6

2x2 − 2=

(x− 3)(x + 2)

2(x− 1)(x + 1)

If there is no common factor the vertical asymptotes of arational function are there where the denominator is 0.So, the vertical asymptotes are x = 1 and x = −1.

(ii) Find the horizontal asymptote of R(x) if there is any.

If the degree of the numerator=the degree of the denomina-tor then the horizontal asymptote is the ratio of the leading

coefficients. So, the horizontal asymptote is y =1

2.

12

(iii) Find the x-intercepts of R(x) if there is any.

The x intercepts of a rational function are there where thenumerator is 0. So, the x intercepts are x = 3 and x = −2.

(iv) Find the y-intercept of R(x) if there is any. Substitute x = 0

(if you can) and solve it for y.So, the y-intercept is y = −6

−2= 3.

(b) Consider the rational function R(x) =x + 3

x2 + x− 2.

(i) Find the vertical asymptotes of R(x) if there is any. Factorthe numerator and the denominator if you can.

x + 3

x2 + x− 2=

x + 3

(x− 1)(x + 2)

If there is no common factor the vertical asymptotes of arational function are there where the denominator is 0.So, the vertical asymptotes are x = 1 and x = −2.

(ii) Find the horizontal asymptote of R(x) if there is any.

If the degree of the numerator<the degree of the denominatorthen the horizontal asymptote is always 0. So, the horizontalasymptote is y = 0.

(iii) Find the x-intercepts of R(x) if there is any.

The x intercepts of a rational function are there where thenumerator is 0. So, the x intercept is x = −3.

(iv) Find the y-intercept of R(x) if there is any.

Substitute x = 0 (if you can) and solve it for y.

So, the y-intercept is y = −3

2.

(c) Consider the rational function R(x) =x3 − x

x2 + 1.

(i) Find the vertical asymptotes of R(x) if there is any. Factorthe numerator and the denominator.

x3 − x

x2 + 1=

x(x− 1)(x + 1)

x2 + 1

If there is no common factor the vertical asymptotes of arational function are there where the denominator is 0. Sincethere is no real solution for x2 + 1 = 0, there is no verticalasymptote.

(ii) Find the horizontal asymptote of R(x) if there is any.

If the degree of the numerator>the degree of the denominatorthen there is no horizontal asymptote.

13

(iii) Find the x-intercepts of R(x) if there is any.

The x intercepts of a rational function are there where thenumerator is 0. So, the x intercepts are x = 0, x = 1 andx = −1.

(iv) Find the y-intercept of R(x) if there is any.

Substitute x = 0 (if you can) and solve it for y.So, the y-intercept is y = 0.

(18) The number of deer in a state forest can be modeled using the therational function

N(t) = 4500(6t2 + 1

3t2 + 1)

where t is the time in years after the herd was first introduced.

(a) Approximately how many deer are there 5 years after they wereintroduced into the forest?

Substitute t = 5 into the formula N(t) = 45006t2 + 1

3t2 + 1. Approxi-

mately 8940 deer.

(b) In the long run, how many deer will there be in this state forest?Explain the reasoning behind your answer. Also explain what pos-sible real-world factors might limit the number of deer in the forest.

You have to find the horizontal asymptote of the rational function

N(t) = 45006t2 + 1

3t2 + 1. The rational function approaches to the hori-

zontal asymptote as t increases. As t →∞ N(t) → 4500 · 63

= 9000deer. The answer is 9000 deer.

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(19) Solve for x.

(a) ln x + ln(x− 15) = ln 34.

ln(x(x− 15)) = ln 34

x(x− 15) = 34

x2 − 15x− 34 = 0

(x + 2)(x− 17) = 0

Possible solutions are x = −2 and x = 17. Checking the possibleanswers, we find that the only solution is x = 17.

(b) log x + log(x + 3) = 1.

log(x(x + 3)) = 1

The equivalent exponential form is

(x(x + 3)) = 10

x2 + 3x− 10 = 0

(x + 5)(x− 2) = 0

So, the solution is x = 2.

(c) e2x − 3ex − 10 = 0.

Substitute y = ex. Then the equation becomes

y2 − 3y − 10 = 0

(y − 5)(y + 2) = 0

So, y = ex = 5 or y = ex = −2. The solution for y = ex = 5 isx = ln 5. There is no solution of y = ex = −2.So, the solution is x = ln 5.

15

(20) (a) Use logarithms to find the solution correct to 3 decimal places.

22x−3 = 32x+1

Take the natural logarithm (ln) of both sides of the equation.

ln 22x−3 = ln 32x+1

Using the 3rd law of logarithm we obtain

(2x− 3) ln 2 = (2x + 1) ln 3

ln 2 = 0.693 and ln 3 = 1.099 with 3 decimal places accuracy. So,

(2x− 3)0.693 = (2x + 1)1.099

After distribution we have

1.386x− 2.079 = 2.198x + 1.099

Now solve the equation for x

−3.178 = 0.812x

x = −3.914

(b) Use the Laws of Logarithms to rewrite the following expression ina form with no logarithm of a product, quotient or power.

log (x4

√y3

z2)

log (x4

√y3

z2) = log (x4y3/2

z) = log (x4)+log (y3/2)−log z = 4 log x+3/2 log y−log z

So, the answer is 4 log x + 3/2 log y − log z.

(c) Rewrite the following expression as a single logarithm.

2 log (x + 1)− 3 log (y − 5) + log z

2 log (x + 1)−3 log (y − 5)+log z = log (x + 1)2−log (y − 5)3+log z = log[(x + 1)2 · z(y − 5)3

]

The answer is log[(x + 1)2 · z(y − 5)3

].

16

(21) Find the time required for an investment of 2500 dollars to grow to 9000dollars at an interest rate of 6.5 percent per year, compounded quarterly.

A(t) = P (1 + r/n)nt

P=$ 2500A(t)=$ 9000r = 0.065n = 4Find t (number of years)

9000 = 2500(1 + 0.065/4)4t

dividing both sides of the equation by 2500

3.6 = (1 + 0.065/4)4t

3.6 = (1.01625)4t

taking the natural logarithm (ln) of both sides of the equation

ln 3.6 = ln((1.01625)4t)

using the 3rd Law of the Logarithm

ln 3.6 = (4t) ln(1.01625)

t =ln 3.6

4 ln(1.01625)= 19.86636111 years

(22) You are in a group of city planners that is trying to determine whetheror not to expand your water supply facilities. To aid in your decision,you will use the information from the last two census figures for the city,showing a population of 80,000 at the start of 1990 and a populationof 88,300 at the start of 2000. Knowing that you can currently supplyenough water for 95,000 people use the exponential model Q(t) = Q0 ·ekt

to determine during what year you will no longer have enough water tomeet the needs of your city.

When t = 0 then Q0 = 80, 000When t = 10 then Q(10) = 88, 330

88, 330 = 80, 000ek10

1.104125 = ek10

ln 1.104125 = k10

ln 1.104125

10= 0.009905316608 = k

17

The exponential model is

Q(t) = 80, 000 · e0.009905316608t

Find t when Q(t) = 95, 000

95, 000 = 80, 000 · e0.009905316608t

1.1875 = e0.009905316608t

ln 1.1875 = 0.009905316608t

ln 1.1875

0.009905316608= 17.349 = t

So, 1990+17=2007, when the city no longer has enough water.

(23) Find the half-life of a radioactive substance if 200 grams of the substancedecays to 180 grams in 2 year.

Q(t) = Q0 · e−kt

Q0 = 200, when t = 2 Q(2) = 180. Find k.

180 = 200 · e−k2

0.9 = e−k2

ln 0.9 = −k2

ln 0.9

−2= 0.05268025783 = k

Thus, the equation describing the procedure is

Q(t) = 200 · e−0.05268025783t

To find the half-life, substitute Q(t) = 100 and calculate t (the timeduring the radio active material loses half of its original mass).

100 = 200 · e−0.05268025783t

1/2 = e−0.05268025783t

ln 1/2 = −0.05268025783t

ln 1/2

−0.05268025783= 13.16 = t is the half-life.

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(24) (a) Find the exponential function f(x) = ax whose graph goes throughthe point. (2, 10.3).

Substitute x = 2 and y = 10.3 and solve the equation for a.

10.3 = a2

Note, the the equation above is not exponential. It is just a simplequadratic equation. Taking the square root of both side of theequation we get

a =√

(10.3) = 3.209361307

So, the solution is

y = 3.209361307x.

(b) For the logarithmic function log3 (x− 4) = y find the domain andthe x-intercept.

To find the domain of log3 (x− 4) = y, note that we can take thelogarithm only positive numbers. So, x − 4 > 0 that is x > 4. Sothe domain is (4,∞).

To find the x intercept substitute y = 0 and solve the equationlog3 (x− 4) = 0. The equivalent exponential form is x−4 = 30 = 1,that is x=5. So, the x intercept is x = 5.

(25) Find all the solutions of the following system of equations.(a)

8x− 3y = −3

5x− 2y = −1

We use the elimination method.

8x− 3y = −3 multiply by -2

5x− 2y = −1 multiply by 3

Then, we get

−16x + 6y = 6

15x− 6y = −3

Add the equations above. Then you eliminate the variable y.

−x = 3

x = −3

19

Substitute x = −3 either of the original equations and solve it fory. We obtain y = −7.So, the solution is x = −3 and y = −7.

(b)x− 2y = 2

y2 − x2 = 2x + 4

We use the substitution method. Let express x from the first equa-tion.

x = 2 + 2y

Substitute into the second equation

y2 − (2 + 2y)2 = 2(2 + 2y) + 4

Combining and rearranging the terms we obtain

3y2 + 12y + 12 = 0

Simplifying by 3

y2 + 4y + 4 = 0

which is

(y + 2)2 = 0

So, y = −2. From the first equation x = 2 + 2y, thus x = −2.The solution is x = −2 and y = −2.

(c)x + 4y = 8

3x + 12y = 2

We use the elimination method.

x + 4y = 8 multiply by -3

3x + 12y = 2

Then, we get

−3x− 12y = −24

3x + 12y = 2

Add the equations above. We obtain,

0 = −22

20

which is impossible. It means there is no solution of this system ofequations. Geometrically it means that the two lines representedby the equations are parallel, so they never intersect.

(d)3x2 + 4y = 17

2x2 + 5y = 2

We use elimination method.

3x2 + 4y = 17 multiply by -2

2x2 + 5y = 2 multiply by 3

We obtain,

−6x2 − 8y = −34

6x2 + 15y = 6

Add the equations above. Then we have,

7y = −28

y = −4

The substitute y = −4 into the first equation. We have

x2 = 11

So x =√

11 and x = −√11.The solutions are x =

√11 and y = −4 or x = −√11 and y = −4

(26) The 12th term of an arithmetic sequence is 32 and the 5th term is 18.Find the difference d and the 20th term.

Using the n-th term formula an = a1+(n−1)d for arithmetic sequence,we have

32 = a12 = a1 + 11d

18 = a5 = a1 + 4d

Subtract the second equation from the first one

14 = 7d

d = 2

Then substitute d = 2 into either of the equations to get a1. So, a1 =10. From the n-th term formula an = a1 +(n−1)d, a20 = 10+9 ·2 = 38.

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(27) An architect designs a theater with 15 seats in the first row, and 18 inthe second , 21 in the third , and so on. If the theater is to have a seatingcapacity of 870 , how many rows must the architect use in his design.

The numbers of seats in the rows form an arithmetic sequence witha1 = 15 and d = 3. The question is how many terms of the arithmeticsequence 15, 18, 21, . . . must be added to get 870. That is, we are askedto find n, the number of rows when Sn = 870. Substituting a1 = 15 ,

d = 3 and Sn = 870 into the formula Sn =n

2[2a1 + (n− 1)d] for the

partial sum of an arithmetic sequence, we get

870 =n

2[2 · 15 + (n− 1)3]

870 = 15 · n +n

2(n− 1)3

Multiply by 2 to get

1740 = 30 · n + n(n− 1)3

Distribute and rearrange the term to get

3n2 + 27n− 1740 = 0

We can simplify by 3

n2 + 9n− 580 = 0

(n + 29)(n− 20) = 0

This gives n = −29 and n = 20. Since n is the number of rows ,wemust have n = 20.

(28) The common ratio in a geometric sequence is 2/5 and the 4th term is5/2. Find the 3rd term.

The n-th term of a geometric sequence is given by the formula an =a1 · rn−1. Thus,

a4 = a1 · r3

Substituting r =2

5and a4 =

5

2we get the following equation

5

2= a1 · (2

5)3

5

2= a1 · ( 8

125).

Let solve the equation for a1,

625

16= a1.

22

Using the n-th term formula of a geometric sequence for n = 3, wehave

a3 = a1 · r2.

Let us substitute625

16= a1 and r =

2

5, then we have

a3 =625

16· (2

5)2

a3 =625

16· ( 4

25)

a3 =25

4

(29) Find the sum.

1 + 3 + 9 + 27 + . . . + 2187

The above sequence is a geometric sequence with ratio r = 3 and firstterm a1 = 1. We have to find out the number of terms n of the givengeometric sequence. The n-th term of a geometric sequence is given bythe formula an = a1 · rn−1. Substitute 2187 = an, r = 3 and a1 = 1 andsolve the equation for n.

2187 = 3n−1

37 = 3n−1

7 = n− 1

8 = n

To find the sum of the first 8 terms of a geometric sequence with

a1 = 1 and r = 3. Using the formula for Sn = a1 · 1− rn

1− rwith n = 8 we

get

S8 = 1 · 1− 38

1− 3

S8 = 3280

23

(30) Determine whether the following sequence arithmetic or geometric andfind the 8th term of the sequence.

27

8,−9

4,3

2,−1

It is a geometric sequence with ratio r = −2

3. Using n-th term formula

for the geometric sequence an = a1 · rn−1 with a1 =27

8, r = −2

3and

n = 8 we get,

a8 =27

8· (−2

3)7

= −16

81.