Algebra 3 Quiz: Geometric Sequences And Series

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Geometric Sequences And SeriesQuestion 1 of 20

In a geometric sequence, $a_2=6$ and $a_5=48$ . What is $a_7$ ?

192

96

384

48

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Algebra 3 Quiz

Algebra 3 Quiz: Geometric Sequences And Series

Practice Geometric Sequences And Series in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Geometric Sequences And Series, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In a geometric sequence, $a_2=6$ and $a_5=48$ . What is $a_7$ ?

$192$ (correct answer)
$96$
$384$
$48$

Explanation: Whenever you see a geometric sequence, isolate the common ratio

r

first. Between

a_2=6

and

a_5=48

, there are 3 steps, so

6r^3=48 \quad\Rightarrow\quad r^3=8 \quad\Rightarrow\quad r=2.

Now

a_7

is 2 steps after

a_5

, so multiply by

r^2

a_7=48\cdot 2^2=48\cdot4=192.

That is the correct value. The choice

96

is one step too soon — it is actually

a_6=48\cdot2

, not

a_7

. The choice

384

comes from multiplying by

2^3=8

instead of

2^2=4

; that would land on

a_8

, or it happens if you count terms from

a_5

a_7

inclusively rather than counting the gaps between them. The choice

48

simply repeats

a_5

, ignoring the common ratio entirely. Remember: in a geometric sequence, the exponent is the number of gaps, not the number of terms listed. To go from

a_5

a_7

is 2 gaps, so use

r^2

. Always confirm how many steps you are moving before multiplying.

Question 2

Find the sum of the first 8 terms of the geometric sequence $3,\;-6,\;12,\;-24,\;\ldots$ .

$-255$ (correct answer)
$255$
$129$
$-129$

Explanation: Whenever you see a geometric sequence, first identify the common ratio by dividing any term by the previous one. Here

-6/3=-2

, so each term is multiplied by

-2

. The sum of the first

n

terms is

S_n=a_1\frac{1-r^n}{1-r}.

With

a_1=3

r=-2

, and

n=8

S_8=3\frac{1-(-2)^8}{1-(-2)} =3\frac{1-256}{3} =-255.

So the 8-term sum is

-255

. Notice

(-2)^8

is positive

256

, not

-128

; using the 7th power gives the sum of only 7 terms, which is

129

. That is the trap behind the

129

choice: it stops one term too soon. The choice

255

has the right magnitude but the wrong sign; because the 8th term is

-384

, adding it to the 7-term sum

129

gives

-255

, not

255

. The choice

-129

is likely the result of taking the 7-term sum

129

and flipping its sign, or mixing up the formula's numerator and denominator. To stay safe, use the formula with

n=8

, and check whether an even number of alternating terms should make the total negative.

Question 3

The repeating decimal $0.272727\ldots$ can be written as an infinite geometric series. Which fraction equals this decimal?

$\frac{27}{100}$
$\frac{3}{11}$ (correct answer)
$\frac{5}{18}$
$\frac{1}{4}$

Explanation: Whenever you see a repeating decimal, think of it as an infinite geometric series whose common ratio is a power of

\frac{1}{10}

. Here,

0.272727\ldots = 0.27 + 0.0027 + 0.000027 + \cdots

This is a geometric series with first term

a=\frac{27}{100}

and common ratio

r=\frac{1}{100}

. Its sum is

\frac{a}{1-r}=\frac{\frac{27}{100}}{1-\frac{1}{100}} =\frac{\frac{27}{100}}{\frac{99}{100}} =\frac{27}{99} =\frac{3}{11}.

\frac{3}{11}

is the correct fraction. The choice

\frac{27}{100}

is just the first two decimal places, not the full repeating value — it equals

0.27

, not

0.272727\ldots

. The choice

\frac{5}{18}

equals

0.2777\ldots

, which repeats a single

7

, not the two-digit block

27

. And

\frac{1}{4}

equals

0.25

, which is close but definitely not the repeating pattern. A quick study tip: for a decimal with a repeating block of length

n

, write the block over

n

nines and simplify. Since

27

repeats,

\frac{27}{99}=\frac{3}{11}

. This shortcut saves time and helps you check your geometric-series work.

Question 4

A ball is dropped from a height of 8 meters. On each bounce it rises to $60\%$ of the height from which it just fell. What total vertical distance does the ball travel before coming to rest?

$12.8$ m
$20$ m
$24$ m
$32$ m (correct answer)

Explanation: Whenever you see a bouncing ball or any repeated-motion distance problem, think geometric series. The key is to track every segment: the ball drops, then rises and falls repeatedly until the bounces become negligible. The initial drop is

8

meters. After that, each bounce reaches

60\%

of the previous height, so the rebound heights are

8(0.6), 8(0.6)^2, 8(0.6)^3, \dots

. For each rebound height, the ball travels up that distance and then down that same distance, so each contributes twice its height. The total distance is therefore

8 + 2\left(8(0.6) + 8(0.6)^2 + 8(0.6)^3 + \cdots\right).

The infinite sum in parentheses is a geometric series with first term

8(0.6)=4.8

and ratio

0.6

. Its sum is

\frac{4.8}{1-0.6}=12

. Twice that is

24

, and adding the initial

8

gives

8 + 24 = 32\text{ m}.

32

m is correct. The wrong answers show common traps.

12.8

m is just the first drop plus the first rise, ignoring all later bounces.

20

m is the sum of only the falling distances

8+4.8+2.88+\cdots

, forgetting that every upward bounce also counts.

24

m correctly sums all the up-and-down bounce distances after the first drop but forgets to include the initial

8

-meter drop. Your takeaway: physically list out the motion first — down, up, down, up — then form a geometric series. Include the initial drop, and remember to double every rebound because each bounce has both an up and a down segment.

Question 5

If $x-2,\;2x+1,\;4x+5$ are consecutive terms of a geometric sequence, what is the value of $x$ ?

$-\frac{7}{11}$
$-1$
$-\frac{11}{3}$
$-\frac{11}{7}$ (correct answer)

Explanation: When you see consecutive terms of a geometric sequence, remember that the ratio between each pair of consecutive terms is constant. So here,

\frac{2x+1}{x-2} = \frac{4x+5}{2x+1}

. Cross-multiplying gives

(2x+1)^2 = (x-2)(4x+5)

. Expand carefully:

4x^2+4x+1 = 4x^2-3x-10

. Cancel the

4x^2

terms:

4x+1 = -3x-10

, so

7x = -11

, and

x = -\frac{11}{7}

. This checks: the terms become

-\frac{25}{7}, -\frac{15}{7}, -\frac{9}{7}

, each multiplied by

\frac35

. The other choices come from common traps.

-\frac{11}{3}

is what you get if you expand

(2x+1)^2

incorrectly as

4x^2+1

, dropping the middle term

4x

-1

makes the sequence arithmetic, not geometric:

-3, -1, 1

have a common difference, not a common ratio.

-\frac{7}{11}

is the reciprocal of the correct answer, suggesting the equation

7x=-11

was solved by flipping the fraction instead of dividing by 7. On exam day, set up the ratio equality first, then expand carefully—especially when squaring binomials. A quick check of the three terms with your answer confirms whether the common ratio is constant.

Question 6

The $n$ th partial sum of a geometric series is $S_n=3(2^n-1)$ . What is the fourth term $a_4$ ?

$48$
$45$
$24$ (correct answer)
$21$

Explanation: When you are given a formula for the partial sum

S_n

, you are being tested on the relationship between a running total and a single term. The key move is to remember that the

n

th term is the difference between consecutive partial sums:

a_n = S_n - S_{n-1}

For this geometric series,

S_4 = 3(2^4-1)=3(16-1)=45

and

S_3 = 3(2^3-1)=3(8-1)=21

So the fourth term is

a_4 = S_4 - S_3 = 45 - 21 = 24.

The choice

45

is the trap of reporting the partial sum

S_4

itself, not the fourth term. The choice

21

S_3

, the sum of the first three terms, not the fourth term. The choice

48

comes from incorrectly using

a_4 = 3(2^4)=48

, treating the exponent as

n

instead of

n-1

for a term; even in the sum formula, the exponent inside is

n

, but the individual term is

3 \cdot 2^{3}

. A quick study tip: whenever a question gives you partial sums, immediately write

a_n = S_n - S_{n-1}

. That single formula converts a potentially tricky series question into a simple subtraction problem.

Question 7

An infinite geometric series has sum $12$ and common ratio $-\frac13$ . What is its second term?

$-4$
$\frac{16}{9}$
$-\frac{16}{3}$ (correct answer)
$16$

Explanation: Whenever an infinite geometric series gives you its sum and common ratio, your first move is to recover the first term using the formula

S = \frac{a}{1-r}

. Here

S=12

and

r=-\frac13

, so

a = S(1-r) = 12\left(1+\frac13\right)=16.

That

16

is the first term, not the answer. The second term is the first term multiplied by the common ratio:

ar = 16 \cdot \left(-\frac13\right) = -\frac{16}{3}.

Now see why the other choices miss the mark.

-4

comes from multiplying the sum

12

by the ratio

-\frac13

; that conflates the total sum with the first term.

\frac{16}{9}

is actually the third term,

ar^2 = 16 \cdot \frac19

; it confuses the term position with the exponent and also drops the negative sign.

16

is the first term you computed correctly, but the question asks for the second term, not the first. On exam day, write the sum formula first, solve for

a

, then take one more step to

ar

. Remember: the infinite sum formula gives you the first term, but the question may ask for a different term. Always check the requested term number and use

ar^{n-1}

Question 8

Which expression gives the $n$ th term of the geometric sequence $\frac12,\;-1,\;2,\;-4,\;\ldots$ ?

$a_n=-\frac12(-2)^{n-1}$
$a_n=\frac12(-2)^{n-1}$ (correct answer)
$a_n=\frac12(-2)^n$
$a_n=(-2)^{n-1}$

Explanation: Whenever you see a geometric sequence, your first move is to identify the first term

a_1

and the common ratio

r

. The

n

th term is always

a_n = a_1 r^{n-1}

. Here the first term is

\frac12

, and dividing consecutive terms gives

r=-2

, since

-1 \div \frac12 = -2

2 \div -1 = -2

, and

-4 \div 2 = -2

. So the expression must be

a_n=\frac12(-2)^{n-1}

. The choice

a_n=\frac12(-2)^{n-1}

is correct because when

n=1

, it gives

\frac12(-2)^0=\frac12

, and then each next term multiplies by

-2

, producing

-1, 2, -4

, and so on. Now look at the distractors. The expression

a_n=-\frac12(-2)^{n-1}

has the wrong sign on the first term: at

n=1

it gives

-\frac12

, which is not the first term. The expression

a_n=\frac12(-2)^n

uses an exponent of

n

instead of

n-1

; at

n=1

it gives

\frac12(-2)=-1

, which is actually the second term, so it is shifted one position too late. Finally,

a_n=(-2)^{n-1}

leaves out the initial factor

\frac12

; at

n=1

it gives

1

, not

\frac12

. A quick study tip: always check your formula by plugging in

n=1

. The exponent should be

n-1

so the first term comes out exactly as given, and don't forget the original starting value.

Question 9

How many terms are in the geometric sequence $3,\;6,\;12,\;\ldots,\;768$ ?

$8$
$9$ (correct answer)
$10$
$7$

Explanation: Whenever you see a geometric sequence, start with the general term formula:

a_n = a_1 r^{n-1}

. Here

a_1 = 3

, and each term doubles, so

r=2

. You need the term number

n

such that

a_n = 768

. Set up the equation:

3(2^{n-1}) = 768

Divide by 3:

2^{n-1} = 256

Since

256 = 2^8

, you get

n-1 = 8

, so

n = 9

. The sequence is

3, 6, 12, 24, 48, 96, 192, 384, 768

, which has exactly 9 terms. The choice

8

is the exponent

n-1

— it counts the number of doublings after the first term, not the number of terms. The choice

7

likely comes from counting only the terms strictly between

3

and

768

, leaving out both endpoints. The choice

10

comes from adding an extra term, such as continuing to

1536

or counting the first term twice. The key habit: in a geometric sequence, the exponent on

r

is always one less than the term number. So when you solve and get an exponent of 8, remember to add 1 back to find 9 terms. That off-by-one trap is exactly what this question tests.

Question 10

What is the sum of the first five terms of the geometric sequence $1,\;-\frac{2}{3},\;\frac{4}{9},\;-\frac{8}{27},\;\ldots$ ?

$\frac{3}{5}$
$\frac{13}{27}$
$\frac{55}{81}$ (correct answer)
$\frac{211}{81}$

Explanation: Whenever you see a geometric sequence, start by finding the common ratio

r

by dividing any term by the one before it. Here,

r=\frac{-2/3}{1}=-\frac{2}{3}.

The sequence alternates signs, and the first five terms are

1,\;-\frac23,\;\frac49,\;-\frac8{27},\;\frac{16}{81}

. You can either add them directly using denominator

81

\frac{81-54+36-24+16}{81}=\frac{55}{81},

or use the finite geometric sum formula

S_5=\frac{1\left(1-(-2/3)^5\right)}{1-(-2/3)} =\frac{1+\frac{32}{243}}{\frac53} =\frac{275}{243}\cdot\frac35 =\frac{55}{81}.

So the sum is

\frac{55}{81}

. As for the other answers,

\frac35

is the sum of the infinite geometric series, since

1/(1+\frac23)=\frac35

; but the question asks for the first five terms only.

\frac{13}{27}

is the sum of only the first four terms, so it is the result of stopping one term too early.

\frac{211}{81}

comes from adding the terms as if all signs were positive, ignoring the fact that the common ratio is negative. In short: always include the sign of

r

, and make sure you use the exact number of terms requested.

Question 11

The second term of an infinite geometric series is $-15$ and the third term is $7.5$ . What is the sum of the series?

$20$ (correct answer)
$-60$
$-10$
$5$

Explanation: Whenever you see an infinite geometric series, the key is to find the common ratio first, then the first term. Here, the third term divided by the second term gives the ratio:

r=\frac{7.5}{-15}=-0.5

Since the second term is

ar=-15

, we have

a(-0.5)=-15 \quad\Rightarrow\quad a=30

Now apply the infinite sum formula:

S=\frac{a}{1-r}=\frac{30}{1-(-0.5)}=\frac{30}{1.5}=20

So the sum is indeed

20

. Why the others miss the mark:

$-60$ comes from using $r=0.5$ instead of $-0.5$ , which ignores the alternating sign. That would give $a=-30$ and sum $-60$ .
$-10$ treats the second term $-15$ as if it were the first term: $-15/(1-(-0.5))=-10$ . But you must divide by the ratio to recover the actual first term.
$5$ treats the third term $7.5$ as the first term: $7.5/(1-(-0.5))=5$ . This skips the step of finding $a$ entirely.

A good habit: always compute

r

first, then use a given term to find

a

before applying

S=\frac{a}{1-r}

. Also watch the sign of

r

— a negative ratio means the series alternates, and ignoring that sign is a common trap.

Question 12

For which values of $x$ does the infinite series $\sum_{n=1}^{\infty}\left(\frac{3}{x-1}\right)^{n-1}$ converge, and what is its sum?

$-4<x<2$ ; $\frac{x-1}{x-4}$
$x<-2$ or $x>4$ ; $\frac{x-1}{x+4}$
$-2<x<4$ ; $\frac{x-1}{x-4}$
$x<-2$ or $x>4$ ; $\frac{x-1}{x-4}$ (correct answer)

Explanation: Whenever you see an infinite series whose terms are powers of a fixed expression, think geometric series: it converges exactly when

|r|<1

, and its sum is

\frac{a}{1-r}

. Here the first term is

\left(\frac{3}{x-1}\right)^0=1

, and the common ratio is

r=\frac{3}{x-1}

. So convergence requires

\left|\frac{3}{x-1}\right|<1 \quad \Rightarrow \quad |x-1|>3.

This means

x-1>3

x-1<-3

, giving the interval

x<-2

x>4

. For those

x

, the sum is

\frac{1}{1-\frac{3}{x-1}}=\frac{x-1}{x-4}.

So the correct choice is the one with interval

x<-2

x>4

and sum

\frac{x-1}{x-4}

. The choice with interval

-4<x<2

misplaces the center of the inequality: the condition is about distance from

1

, not from

-1

, and it must be greater than

3

. The choice with sum

\frac{x-1}{x+4}

uses the correct interval but adds the ratio in the denominator instead of subtracting it:

1-r

, not

1+r

. The choice

-2<x<4

reverses the inequality; that interval actually makes

|r|>1

, so the series diverges there. Study tip: for any geometric series, write down

a

and

r

first. Then solve

|r|<1

carefully, and simplify

\frac{a}{1-r}

separately. That keeps the interval and the sum from getting tangled together.

Question 13

An infinite geometric series has sum $12$ and common ratio $-\frac{1}{3}$ . What is the first term?

$-16$
$-\frac{16}{3}$
$-4$
$16$ (correct answer)

Explanation: Since

S_\infty=\frac{a_1}{1-r}

, we have

12=\frac{a_1}{1-(-1/3)}=\frac{a_1}{4/3}

, so

a_1=16

. The value

-16

has the wrong sign;

-\frac{16}{3}

is actually the second term,

16(-1/3)=-16/3

;

-4

comes from computing

S_\infty r=12(-1/3)=-4

Question 14

For what values of $x$ does the infinite geometric series $\displaystyle \sum_{n=0}^{\infty}(2x-1)^n$ converge?

$0\le x\le1$
$x<1$
$0\le x<1$
$0<x<1$ (correct answer)

Explanation: The ratio is

2x-1

. A geometric series converges only when

|2x-1|<1

, so

-1<2x-1<1

, which gives

0<x<1

. The interval

0\le x\le1

includes endpoints where

|r|=1

;

x<1

omits the lower bound;

0\le x<1

incorrectly includes

x=0

, where

r=-1

Question 15

A geometric sequence has first term $5$ , common ratio $2$ , and last term $160$ . How many terms are in the sequence?

$5$
$6$ (correct answer)
$7$
$8$

Explanation: The last term satisfies

5(2)^{n-1}=160

, so

2^{n-1}=32=2^5

. Thus

n-1=5

, so

n=6

. The value 5 comes from forgetting to add 1 after solving the exponent; 7 and8 come from off-by-one errors in the exponent.

Question 16

In a geometric sequence, $a_3=12$ and $a_6=96$ . What is the sum of the first $8$ terms?

$381$
$765$ (correct answer)
$1530$
$3060$

Explanation: Since

a_6/a_3=r^3=8

r=2

. Then

a_1=a_3/r^2=12/4=3

, so

S_8=3(2^8-1)=3(255)=765

. The value

381

S_7

, using one fewer term;

1530

comes from using

a_1=6

instead of 3;

3060

comes from using

a_1=12

instead of 3.

Question 17

In a geometric sequence, $a_4=40$ and $a_6=160$ . If the sum of the first $n$ terms is $2555$ , what is $n$ ?

$8$
$9$ (correct answer)
$10$
$11$

Explanation: Since

a_6/a_4=r^2=4

r=\pm 2

. If

r=-2

, then

a_1=-5

and

S_n=-\frac{5}{3}(1-(-2)^n)

; odd

n

gives a negative sum, while even

n

would require

2^n=1534

, impossible. Thus

r=2

, so

a_1=5

, and

5(2^n-1)=2555

. Therefore

2^n=512

, so

n=9

. The value 10 may come from solving

2^{n-1}=512

; 8 and 11 are common index/exponent slips.

Question 18

In a geometric sequence, $a_4=2$ and $a_8=162$ . What is $a_6$ ?

$-18$
$18$ (correct answer)
$54$
$162$

Explanation: Since

\frac{a_8}{a_4}=r^4=\frac{162}{2}=81

, we get

r=\pm3

. But

a_6=a_4r^2=2\cdot9=18

, so the sign of

r

does not affect the sixth term. The value

-18

incorrectly assumes that a negative ratio makes

a_6

negative;

54

a_7

when the positive ratio is used;

162

a_8

Question 19

A geometric series has first term $48$ and common ratio $\frac{1}{2}$ . What is the least number of terms for which the sum exceeds $90$ ?

$3$
$4$
$5$ (correct answer)
$6$

Explanation: The partial sum is

S_n=48\frac{1-(1/2)^n}{1-1/2}=96\left(1-2^{-n}\right)

. For

n=4

S_4=96(1-1/16)=90

, which does not exceed 90. For

n=5

S_5=96(1-1/32)=93

, so the least value is 5. The value 4 corresponds to reaching exactly 90 rather than exceeding it.

Question 20

For a geometric series, the sum of the first $2$ terms is $8$ , the sum of the first $4$ terms is $80$ , and the first term is positive. What is the common ratio?

$3$ (correct answer)
$-3$
$10$
$3$ or $-3$

Explanation: Let

a

be the first term and

r

the ratio. Then

a(1+r)=8

and

a(1+r+r^2+r^3)=80

. Dividing gives

1+r^2=10

, so

r^2=9

, meaning

r=\pm3

. If

r=-3

, then

a=8/(1-3)=-4

, contradicting the positivity of the first term, so

r=3

. The value

-3

ignores that positivity condition;

10

comes from treating

80/8

as the ratio;

3

-3

forgets to check the first-term sign.