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Geometric Series: 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 The alternating terms

A geometric sequence begins $2,-8,32,\ldots$ , each term being $-4$ times the one before it. Find the sum of its first five terms.
1. Hint 1
  A finite geometric sum needs only the first term, the ratio, and the number of terms.
2. Hint 2
  Substitute $r$ with its minus sign attached, both in the power and in the denominator.
3. Hint 3
  An odd power of a negative ratio is negative, so subtracting it adds to the numerator.
Answer

$410$ .

Full solution

Read off the three inputs: the first term is $a_1=2$ , the ratio is $r=-4$ , and five terms are being added.
Since $r\ne1$ , the finite-sum formula applies:
$S_5=\frac{2\left(1-(-4)^{5}\right)}{1-(-4)}.$
The fifth power of a negative number is negative, so $(-4)^{5}=-1024$ and the numerator is
$2(1+1024)=2050.$
The denominator is $1-(-4)=5$ , so
$S_5=\frac{2050}{5}=410.$
Adding the five terms directly, $2-8+32-128+512=410$ , confirms it.

Answer

$410$ .

Key idea

The finite geometric sum formula handles a negative ratio as long as the sign travels into every power.
Problem 1: Mark complete
Problem 2 The sigma total

Decide whether the infinite series $\sum_{k=1}^{\infty}8\left(-\frac14\right)^{k-1}$ has a finite sum, and if it does, find it.
1. Hint 1
  Read the first term and the ratio straight off the summand before deciding anything.
2. Hint 2
  An infinite geometric series has a finite sum exactly when the ratio satisfies $|r|<1$ .
3. Hint 3
  Carry the minus sign into $S=\frac{a_1}{1-r}$ , so the denominator grows rather than shrinks.
Answer

It does have a finite sum: $S=\frac{32}{5}$ , or $6.4$ .

Full solution

At $k=1$ the summand is $8\left(-\tfrac14\right)^{0}$ , which is $8$ , so the first term is $a_1=8$ .
Each step raises the exponent by one, so every term is $-\tfrac14$ times the one before it, giving
$r=-\tfrac14.$
Because $\left|-\tfrac14\right|=\tfrac14<1$ , the terms shrink toward zero and the series does have a finite sum.

Put the ratio into the infinite-sum formula with its sign attached:
$\begin{aligned} S &= \frac{8}{1-\left(-\tfrac14\right)} \\ &= \frac{8}{\tfrac54} \\ &= \frac{32}{5} \end{aligned}$
As a decimal that is $6.4$ .
The running totals $8$ , $6$ , $6.5$ , $6.375$ swing above and below it by ever smaller amounts, as a negative ratio should.

Answer

It does have a finite sum: $S=\frac{32}{5}$ , or $6.4$ .

Key idea

A sigma with an infinite upper limit is summed by $S=\frac{a_1}{1-r}$ , once the ratio has passed the test $|r|<1$ .
Problem 2: Mark complete
Problem 3 The assembly charges

An assembly has four layers. The material charge for the first layer is 7 dollars, and each later layer has three times the preceding material charge. Each layer also has a fixed handling charge of 2.50 dollars. Find the total charge for all four layers.
1. Hint 1
  Separate material charges from handling charges.
2. Hint 2
  The material charges form a four-term geometric series.
3. Hint 3
  Add the handling charge once for each layer.
Answer

290 dollars.

Full solution

The material charges have first term $7$ , ratio $3$ , and count $4$ .
Their sum is
$M=\frac{7(3^4-1)}{3-1}.$
Since $3^4=81$ , this gives $M=280$ .
The handling total is
$4\cdot2.50=10.$
Therefore the complete charge is
$280+10=290.$
Directly, the material charges $7,21,63,189$ total $280$ , and all four handling charges are included.

Answer

290 dollars.

Key idea

An added fixed charge must be totaled separately from charges that grow geometrically.
Problem 3: Mark complete
Problem 4 The decimal prefix

Write $0.2\overline{14}=0.2141414\ldots$ as a fraction in lowest terms.
1. Hint 1
  Separate the nonrepeating tenths digit from the repeating part.
2. Hint 2
  The repeating blocks begin with $14/1000$ and have ratio $1/100$ .
3. Hint 3
  Sum that infinite series, then add the nonrepeating part.
Answer

$\frac{106}{495}$ .

Full solution

The nonrepeating part is $1/5$ .
The repeated blocks form a geometric series with first term $14/1000$ and ratio $1/100$ .
Since the ratio has magnitude less than $1$ , its sum is
$\frac{14/1000}{1-1/100}=\frac{7}{495}.$
Therefore the decimal equals
$\frac15+\frac{7}{495}=\frac{106}{495}.$
The numerator has no common factor with $495=5\cdot9\cdot11$ , so the fraction is reduced.

Answer

$\frac{106}{495}$ .

Key idea

A nonrepeating decimal prefix is added separately to the geometric series formed by the repeating blocks.
Problem 4: Mark complete
Problem 5 The running display

A display adds the positive entries $12,3,\frac34,\ldots$ , each a quarter of the preceding entry. Find the least number of entries needed for its running total to be within $\frac15$ of the infinite total, meaning that the remaining difference is at most $\frac15$ .
1. Hint 1
  Find the infinite total from the first term and ratio.
2. Hint 2
  Subtract the finite sum from the infinite total.
3. Hint 3
  Compare the remaining differences at neighboring term counts.
Answer

$4$ entries.

Full solution

The infinite total is
$12/(1-1/4)=16.$
The finite total is
$S_n=16\left(1-4^{-n}\right).$
Thus the remaining difference is $16\cdot4^{-n}$ .
At three entries it is
$16\cdot4^{-3}=\frac14,$
which is greater than $1/5$ .
At four entries it is
$16\cdot4^{-4}=\frac{1}{16},$
which is less than $1/5$ .
The positive difference is divided by four at every step, so four is the least possible count.
Directly, the first four entries total $\frac{255}{16}$ , which falls $\frac{1}{16}$ short of $16$ .

Answer

$4$ entries.

Key idea

The difference between a geometric infinite sum and its partial sum gives the exact amount still missing.
Problem 5: Mark complete
Problem 6 The paired totals

An infinite geometric series has real first term $a_1\ne0$ and real ratio $r$ . Its sum is $3$ , and its first two terms have sum $\frac94$ . Find all possible pairs consisting of its first term and common ratio.
1. Hint 1
  Use the infinite total to express the first term in terms of the ratio.
2. Hint 2
  The first two terms have sum $a_1(1+r)$ .
3. Hint 3
  The resulting equation involves $r^2$ , so check both signs.
Answer

$(a_1,r)=(\frac32,\frac12)$ or $(\frac92,-\frac12)$ .

Full solution

An infinite geometric series has a finite sum only when $|r|<1$ , and here
$3=a_1/(1-r).$
Hence
$a_1=3(1-r).$
The two-term condition gives
$3(1-r)(1+r)=\frac94.$
Dividing by $3$ and using the difference of squares gives $1-r^2=3/4$ , so
$r^2=\frac14.$
Thus $r=1/2$ or $r=-1/2$ , both allowed.
They give first terms $3/2$ and $9/2$ , respectively.
Their two-term sums are $3/2+3/4=9/4$ and $9/2-9/4=9/4$ , and their infinite sums both equal $3$ .

Answer

$(a_1,r)=(\frac32,\frac12)$ or $(\frac92,-\frac12)$ .

Key idea

A partial total and an infinite total can leave a sign choice for a geometric ratio.
Problem 6: Mark complete
Problem 7 The signed adjustments

Six adjustments are $10,-5,\frac52,\ldots$ , each $-\frac12$ times the preceding adjustment. Find their net total and the total of their absolute values.
1. Hint 1
  The signed adjustments form one geometric series.
2. Hint 2
  Taking absolute values changes the ratio to its positive magnitude.
3. Hint 3
  Use six terms in each finite sum.
Answer

Net total $\frac{105}{16}$ ; absolute-value total $\frac{315}{16}$ .

Full solution

The signed ratio is $-1/2$ , and its sixth power is $1/64$ .
Since $r=-1/2$ , the denominator $1-r$ equals $1+1/2$ .
The net total is
$N=\frac{10(1-1/64)}{1+1/2}.$
This simplifies to $N=105/16$ .
For absolute values, the ratio is $1/2$ , so
$A=\frac{10(1-1/64)}{1-1/2}.$
This gives $A=315/16$ .
The six absolute values are $10,5,5/2,5/4,5/8,5/16$ .
Over denominator $16$ , their numerators add to $160+80+40+20+10+5=315$ ; alternating the signs gives
$160-80+40-20+10-5=105.$

Answer

Net total $\frac{105}{16}$ ; absolute-value total $\frac{315}{16}$ .

Key idea

Net change uses signed terms, while total adjustment size uses their absolute values.
Problem 7: Mark complete
Problem 8 The middle block

A geometric sequence has ratio $r\ne1$ , and $m$ and $n$ are positions with $m\le n$ . The terms in positions $m$ through $n$ form a geometric series of their own. Write the sum of that block in terms of $a_m$ , $r$ , $m$ and $n$ . Then evaluate the block for the sequence with $a_1=10$ and $r=3$ , taking positions $2$ through $5$ .
1. Hint 1
  A run of consecutive terms is itself geometric, and its own first term sits at the start of the run.
2. Hint 2
  Count how many terms run from position $m$ to position $n$ with both ends included.
3. Hint 3
  Then use the finite-sum formula on that block, or subtract $S_{m-1}$ from $S_n$ .
Answer

The block sum is $\frac{a_m\left(r^{n-m+1}-1\right)}{r-1}$ , equivalently $\frac{a_m\left(1-r^{n-m+1}\right)}{1-r}$ ; for $a_1=10$ , $r=3$ and positions $2$ through $5$ it is $1200$ .

Full solution

Call the block sum $B$ .
Its first term is the one in position $m$ , namely $a_m$ , and every later entry is still $r$ times the entry before it, so the block is a geometric series with ratio $r$ .
Counting positions $m$ through $n$ with both ends included gives $n-m+1$ terms.

Feeding that first term and that count into the finite-sum formula gives
$B=\frac{a_m\left(r^{n-m+1}-1\right)}{r-1}.$
Since $r\ne1$ the denominator is not zero.
The version with $1-r$ underneath is the same number, and subtracting $S_{m-1}$ from $S_n$ gives it too.

Now take $a_1=10$ , $r=3$ , $m=2$ and $n=5$ .
The term in position $2$ is $a_2=10\cdot3^{1}=30$ , and the block holds $5-2+1=4$ terms, so
$\begin{aligned} B &= \frac{30\left(3^{4}-1\right)}{3-1} \\ &= \frac{30\cdot80}{2} \\ &= 1200 \end{aligned}$
Adding the four terms directly, $30+90+270+810$ , gives $1200$ as well.

Answer

The block sum is $\frac{a_m\left(r^{n-m+1}-1\right)}{r-1}$ , equivalently $\frac{a_m\left(1-r^{n-m+1}\right)}{1-r}$ ; for $a_1=10$ , $r=3$ and positions $2$ through $5$ it is $1200$ .

Key idea

A run of consecutive terms of a geometric sequence is itself geometric, so the same sum formula applies with the run's own first term and count.
Problem 8: Mark complete
Problem 9 The ratio built from $x$

For a real number $x$ , an infinite series begins $1+\frac{x}{4}+\left(\frac{x}{4}\right)^{2}+\cdots$ , each term being $\frac{x}{4}$ times the one before it. Find every value of $x$ for which this series has a finite sum, and give that sum in terms of $x$ .
1. Hint 1
  An infinite geometric series settles on a finite total only for ratios of a certain size.
2. Hint 2
  Here the ratio is $\frac{x}{4}$ , so that size condition turns into an inequality in $x$ .
3. Hint 3
  The first term is $1$ ; put the ratio into $S=\frac{a_1}{1-r}$ , then clear the inner fraction.
Answer

$|x|<4$ , that is $-4<x<4$ ; the sum is $S=\frac{4}{4-x}$ , equivalently $\frac{1}{1-x/4}$ .

Full solution

The first term is $1$ and the ratio is
$r=\frac{x}{4}.$
An infinite geometric series has a finite sum exactly when $\left|r\right|<1$ , so the condition here is
$\left|\frac{x}{4}\right|<1.$
Multiplying that inequality by the positive number $4$ leaves $|x|<4$ , which says the same as $-4<x<4$ .

For those values the infinite-sum formula applies with $a_1=1$ :
$S=\frac{1}{1-\frac{x}{4}}.$
Multiplying the top and the bottom by $4$ clears the fraction inside the denominator:
$S=\frac{4}{4-x}.$
The denominator $4-x$ is never zero on $-4<x<4$ , so the sum is defined everywhere the series converges.

Check one value.
At $x=2$ the series is $1+\tfrac12+\tfrac14+\cdots$ , whose total is $2$ , and the formula gives
$\frac{4}{4-2}=2.$
The two ends are excluded for good reason: at $x=4$ every term equals $1$ , and at $x=-4$ the terms flip between $1$ and $-1$ , so neither running total settles.

Answer

$|x|<4$ , that is $-4<x<4$ ; the sum is $S=\frac{4}{4-x}$ , equivalently $\frac{1}{1-x/4}$ .

Key idea

When the ratio carries an unknown, the convergence test $|r|<1$ becomes an inequality that fixes the values the sum formula may be used on.
Problem 9: Mark complete
Problem 10 Lina's running totals

A report adds the first $n$ terms of a geometric sequence whose first term $a_1$ is not zero. It gives total $na_1$ for every positive integer $n$ . Lina says the ratio must be $1$ , so the infinite series has no finite sum. Is she correct? Explain.
1. Hint 1
  The claim must hold at every term count, so a single small count is enough to test it.
2. Hint 2
  Take the report at $n=2$ , and divide by $a_1$ , which you are told is not zero.
3. Hint 3
  Then describe what adding arbitrarily many equal nonzero terms does.
Answer

Yes; $r=1$ , and the infinite series diverges.

Full solution

At $n=2$ , the stated total gives
$a_1+a_1r=2a_1.$
Since $a_1\ne0$ , division by $a_1$ gives $1+r=2$ , hence $r=1$ .
Every term is therefore $a_1$ .
The partial sum $na_1$ grows in magnitude without bound because $a_1$ is nonzero, so it cannot approach a finite total.

Answer

Yes; $r=1$ , and the infinite series diverges.

Key idea

The formula for a constant geometric sum also identifies the ratio and explains its divergent infinite total.
Problem 10: Mark complete

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