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Geometric Sequences: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Neighbors with a shared factor

Two neighboring terms of a geometric sequence are $a_5=-2t$ and $a_6=5t$ , where $t\ne0$ . Find the common ratio.
1. Hint 1
  The common ratio compares the later term with the earlier term.
2. Hint 2
  Divide $5t$ by $-2t$ and cancel the nonzero factor.
Answer

$r=-\frac52$ .

Full solution

Because $t\ne0$ , the earlier term is nonzero.
Dividing the neighboring terms gives
$r=\frac{5t}{-2t}.$
Canceling $t$ gives
$r=-\frac52.$
Multiplying $-2t$ by this ratio returns $5t$ .

Answer

$r=-\frac52$ .

Key idea

A shared nonzero scale factor cancels when neighboring geometric terms are divided.
Problem 1: Mark complete
Problem 2 Four ratios, four behaviors

Four geometric sequences each start at $a_1=6$ , with common ratios $r=-3$ , $r=\frac34$ , $r=1$ and $r=-\frac45$ . Using the phrases grow in size, shrink toward zero, stay constant and alternate in sign, describe what the terms of each sequence do.
1. Hint 1
  The size of a ratio and the sign of a ratio control different features of a sequence.
2. Hint 2
  Compare each ratio with $1$ in size, then look separately at whether it is negative.
Answer

$r=-3$ : grow in size and alternate in sign. $r=\frac34$ : shrink toward zero. $r=1$ : stay constant. $r=-\frac45$ : shrink toward zero and alternate in sign.

Full solution

Every step multiplies the size of a term by $|r|$ , so compare each ratio with $1$ in size.
Since $|-3|=3$ is greater than $1$ , that sequence grows in size, and the negative ratio flips the sign at each step, giving terms $6,-18,54,-162$ and onward.
Since $\frac34$ lies between $0$ and $1$ , those terms shrink toward zero and stay positive.
The ratio $1$ leaves every term equal to $6$ .
Since $\left|-\frac45\right|=\frac45$ is less than $1$ , the last sequence shrinks toward zero as well, while its negative ratio alternates the signs, giving terms $6,-4.8,3.84,-3.072$ and onward.

Answer

$r=-3$ : grow in size and alternate in sign. $r=\frac34$ : shrink toward zero. $r=1$ : stay constant. $r=-\frac45$ : shrink toward zero and alternate in sign.

Key idea

The size of a common ratio decides growth, decay or a steady size, and its sign decides whether the terms alternate.
Problem 2: Mark complete
Problem 3 Every term equal to one

For a real number $q$ , a sequence has $a_1=1$ and $a_{n+1}=(q-4)a_n$ . Find the value of $q$ that makes every term equal to $1$ .
1. Hint 1
  A nonzero term stays unchanged when multiplied by $1$ .
2. Hint 2
  Set the multiplier $q-4$ equal to the required ratio.
Answer

$q=5$ .

Full solution

The common ratio must be $1$ , so
$q-4=1.$
Hence
$q=5.$
With this value, the recurrence multiplies each $1$ by $1$ , producing another $1$ .

Answer

$q=5$ .

Key idea

A nonzero geometric sequence is constant when its common ratio is one.
Problem 3: Mark complete
Problem 4 The panel dimensions

The first panel is $3$ centimeters wide and $16$ centimeters high. Each later panel is twice as wide and half as high as the preceding panel. Find the fifth panel dimensions and decide whether the sequence of panel areas is geometric.
1. Hint 1
  Each dimension changes four times before panel five.
2. Hint 2
  Write a separate geometric rule for width and height.
3. Hint 3
  Multiply the two dimension ratios to find the area ratio.
Answer

Width $48$ centimeters; height $1$ centimeter; areas are geometric with ratio $1$ .

Full solution

Four width changes give
$w_5=3\cdot2^4.$
Thus the width is $48$ centimeters.
Four height changes give
$h_5=16\left(\frac12\right)^4.$
Thus the height is $1$ centimeter.
The area ratio is
$2\cdot\frac12=1.$
Every panel has area $48$ square centimeters, so the areas form a nonzero constant geometric sequence.

Answer

Width $48$ centimeters; height $1$ centimeter; areas are geometric with ratio $1$ .

Key idea

When two dimensions change geometrically, their ratios multiply to give the area ratio.
Problem 4: Mark complete
Problem 5 Every other term

A geometric sequence has $a_n=7(-2)^{n-1}$ for positive integers $n$ . The terms in positions $1,3,5,7,\ldots$ are kept in order and called $b_1,b_2,\ldots$ . Write $b_n$ in the form $Ar^{n-1}$ , find $b_4$ , and state whether the kept terms alternate in sign.
1. Hint 1
  One step of the kept sequence passes over two steps of the original sequence.
2. Hint 2
  The first kept value is $a_1$ , and two original steps multiply by $(-2)^2$ .
3. Hint 3
  The sign of the new ratio settles the question about alternation.
Answer

$b_n=7\cdot4^{n-1}$ and $b_4=448$ ; the kept terms do not alternate in sign.

Full solution

The first kept value is $a_1=7$ .
Two original steps multiply by
$(-2)^2=4.$
Therefore
$b_n=7\cdot4^{n-1}.$
In particular,
$b_4=7\cdot4^3.$
This gives $b_4=448$ .
Its original position is $7$ , and $7(-2)^6=448$ , which checks the result.
The new ratio $4$ is positive, so the kept terms all stay positive and do not alternate, even though the original terms do.

Answer

$b_n=7\cdot4^{n-1}$ and $b_4=448$ ; the kept terms do not alternate in sign.

Key idea

Keeping equally spaced terms raises the original ratio to the spacing, so an even spacing turns a negative ratio into a positive one.
Problem 5: Mark complete
Problem 6 Two conditions, one sequence

A nonzero geometric sequence satisfies $a_4=-27a_1$ and $a_1+a_2=-10$ . Find its common ratio, first term, and sixth term.
1. Hint 1
  The first condition spans three multiplication steps.
2. Hint 2
  Express the second term as $a_1r$ .
3. Hint 3
  Once the ratio and first term are known, count five steps to the sixth term.
Answer

$r=-3$ , $a_1=5$ , and $a_6=-1215$ .

Full solution

Since $a_1\ne0$ , dividing the first condition by $a_1$ gives
$r^3=-27.$
The real ratio is $r=-3$ .
The second term is $a_1r$ , so the second condition reads
$a_1+a_1(-3)=-10.$
Taking out the common factor gives
$a_1(1-3)=-10.$
Thus $a_1=5$ , and
$a_6=5(-3)^5.$
Hence $a_6=-1215$ .
The first two terms are $5$ and $-15$ , totaling $-10$ , while the fourth term is
$-135=-27\cdot5.$

Answer

$r=-3$ , $a_1=5$ , and $a_6=-1215$ .

Key idea

A relation between terms an odd number of steps apart fixes the ratio uniquely, while a second condition fixes the first term.
Problem 6: Mark complete
Problem 7 Does 20 appear?

A sequence has $a_n=8(1.5)^{n-1}$ for positive integers $n$ . Does the value $20$ occur as a term of this sequence? Support your decision with the possible index, rounded to the nearest hundredth.
1. Hint 1
  A value sits in the sequence only when some allowed position produces it.
2. Hint 2
  Set the term formula equal to $20$ and isolate the power of $1.5$ .
3. Hint 3
  Use logarithms, then check whether the possible index is a positive integer.
Answer

No; the possible index is $n\approx3.26$ .

Full solution

A matching position would satisfy
$8(1.5)^{n-1}=20.$
Dividing by the nonzero number $8$ gives
$1.5^{n-1}=2.5.$
Taking logarithms gives
$n=1+\frac{\log(2.5)}{\log(1.5)}.$
The denominator is nonzero because $1.5\ne1$ .
To the nearest hundredth, $n\approx3.26$ .
This is not an integer, so $20$ is not a term.
As a check, $a_3=18$ and $a_4=27$ , so the increasing sequence steps over $20$ .

Answer

No; the possible index is $n\approx3.26$ .

Key idea

A value belongs to a sequence only when its term equation gives an allowed integer position.
Problem 7: Mark complete
Problem 8 Ravi's sixth reading

A tank loses one quarter of its contents each hour, so the hourly readings form a geometric sequence, and the first reading is $512$ liters. To find the sixth reading, Ravi writes $512\left(\frac34\right)^6$ and reports $91.125$ liters. Identify the mistake in his work and give the correct sixth reading.
1. Hint 1
  The exponent counts the multiplications made after the first reading, not the position itself.
2. Hint 2
  Each hour keeps three quarters of the amount, and the sixth reading comes five hours after the first.
3. Hint 3
  Work out the power of $\frac34$ before multiplying by $512$ .
Answer

Ravi used the exponent $6$ instead of $5$ ; the correct sixth reading is $121.5$ liters.

Full solution

Losing one quarter leaves three quarters, so the hourly ratio is
$r=\frac34.$
Reaching the sixth reading from the first takes five hours, so the ratio is applied five times, not six.
The correct value is
$a_6=512\left(\frac34\right)^5.$
Since $\left(\frac34\right)^5=\frac{243}{1024}$ , this gives $a_6=121.5$ liters.
Ravi's exponent $6$ describes the seventh reading, $91.125$ liters, which is one hour too far along.

Answer

Ravi used the exponent $6$ instead of $5$ ; the correct sixth reading is $121.5$ liters.

Key idea

Reaching the $n$ th term of a geometric sequence applies the ratio $n-1$ times, one fewer than the position number.
Problem 8: Mark complete
Problem 9 The matching endpoints

A real geometric sequence has nonzero terms and satisfies $a_7=a_3$ . Noor says the ratio is either $1$ or $-1$ , so every term has the same absolute value. Is she correct? Justify your answer.
1. Hint 1
  The equal terms are four multiplication steps apart.
2. Hint 2
  Divide by the nonzero earlier term.
3. Hint 3
  Find the real numbers whose fourth power equals $1$ .
Answer

Yes; $r=1$ or $r=-1$ , and all terms have equal absolute value.

Full solution

The relation between the positions is
$a_7=a_3r^4.$
Since $a_3\ne0$ , equality of the terms gives
$r^4=1.$
Moving everything to one side makes the left side a difference of squares, so
$(r^2-1)(r^2+1)=0.$
A real $r$ has $r^2\ge0$ , so $r^2+1$ is never zero, leaving
$r^2=1.$
Hence $r=1$ or $r=-1$ .
Both have $|r|=1$ , so multiplication preserves every term magnitude.
The first ratio gives constancy and the second gives alternation of signs.

Answer

Yes; $r=1$ or $r=-1$ , and all terms have equal absolute value.

Key idea

Equal nonzero geometric terms an even number of steps apart can allow either a positive or a negative ratio.
Problem 9: Mark complete
Problem 10 Can a term be zero?

A list of numbers is claimed to be geometric, with $a_2=12$ and $a_5=0$ . Calling it geometric is meant to mean that every term is nonzero and that the same nonzero ratio connects neighboring terms. Can both stated values occur in such a sequence? Explain.
1. Hint 1
  Relate the fifth term to the second through three ratio steps.
2. Hint 2
  A product of nonzero real numbers is nonzero.
3. Hint 3
  Check whether choosing a zero ratio could satisfy the stated definition.
Answer

No; no sequence of that kind has both of these terms.

Full solution

The geometric relation would be
$a_5=12r^3.$
Because $12\ne0$ and $r\ne0$ , this expression is nonzero, contradicting $a_5=0$ .
Choosing $r=0$ would violate the required nonzero ratio and nonzero terms.
Therefore no allowed ratio fits both values.

Answer

No; no sequence of that kind has both of these terms.

Key idea

With a nonzero first term and nonzero common ratio, every geometric term remains nonzero.
Problem 10: Mark complete

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