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Arithmetic 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 The stored entries

An arithmetic sequence has $a_1=-2$ and common difference $d=6$ . Find $a_{13}$ .
1. Hint 1
  Reaching a position costs one fewer equal steps than the position number, because the first term is already there.
2. Hint 2
  Write the explicit rule from the first term and the common difference, then substitute the requested position.
Answer

$a_{13}=70$ .

Full solution

Reaching position $13$ takes $12$ steps of $6$ from the first term, so
$a_{13}=-2+12(6)$
which is $70$ .

Answer

$a_{13}=70$ .

Key idea

The $n$ th term sits $n-1$ equal steps from the first term.
Problem 1: Mark complete
Problem 2 The covered digit

The three numbers $52$ , $4\square$ , and $38$ , in that order, form an arithmetic sequence, where the square hides the ones digit of the middle number. Find the middle number.
1. Hint 1
  The first and last numbers are separated by two equal steps.
2. Hint 2
  Find the signed change per step, then add one such step to the first number.
Answer

$45$ , so the covered digit is $5$ .

Full solution

Two steps together change $52$ to $38$ .
The total change is
$38-52=-14$
so each step changes the value by
$d=\frac{-14}{2}$
which is $-7$ .

The middle number is
$52-7=45$
so the covered digit is $5$ .
Checking, $45-52=-7$ and $38-45=-7$ , so both consecutive differences agree.

Answer

$45$ , so the covered digit is $5$ .

Key idea

Equal steps between known terms can recover a missing term and its digits.
Problem 2: Mark complete
Problem 3 The decimal record

A sequence starts with $a_1=-2.5$ and follows $a_n=a_{n-1}+0.75$ for every integer $n\ge2$ . Write its explicit rule in the form $a_n=pn+q$ for positive integers $n$ .
1. Hint 1
  Count the equal changes needed to reach position $n$ from the first position.
2. Hint 2
  Start with the first value plus $n-1$ copies of the common difference, then distribute.
Answer

$a_n=0.75n-3.25$ for integers $n\ge1$ .

Full solution

The common difference is $0.75$ .
Reaching position $n$ takes $n-1$ additions to the first term.
$a_n=-2.5+0.75(n-1)$

Distribute and combine the constants.
$\begin{aligned}a_n&=-2.5+0.75n-0.75\\a_n&=0.75n-3.25\end{aligned}$

At $n=1$ the rule gives
$0.75-3.25=-2.5.$
Increasing $n$ by one increases the output by $0.75$ , matching both parts of the recursive rule.

Answer

$a_n=0.75n-3.25$ for integers $n\ge1$ .

Key idea

An explicit arithmetic-sequence rule combines the first value with one fewer changes than its position.
Problem 3: Mark complete
Problem 4 The selected entries

An arithmetic sequence has $a_1=4$ and common difference $2.5$ . A device keeps entries in positions $1$ , $4$ , $7$ , $10$ , and so on, skipping two entries between kept entries. It calls the kept sequence $b_1,b_2,b_3,\ldots$ . Give an explicit rule for $b_n$ in the form $b_n=b_1+(n-1)D$ and a recursive rule with its starting value.
1. Hint 1
  One move in the kept sequence passes through several moves in the original sequence.
2. Hint 2
  Find the original position of the $n$ th kept entry and the total change between two kept entries.
Answer

$b_n=4+7.5(n-1)$ for integers $n\ge1$ ; $b_1=4$ , $b_n=b_{n-1}+7.5$ for integers $n\ge2$ .

Full solution

Each kept entry is three original positions after the previous one.
Its change is
$D=3(2.5)$
so $D=7.5$ .
The first kept value is $4$ .

The explicit rule is
$b_n=4+7.5(n-1)$
for positive integers $n$ .
The recursive rule starts at $b_1=4$ and adds $7.5$ each time:
$b_n=b_{n-1}+7.5$
for integers $n\ge2$ .

The first three kept values are $4$ , $11.5$ , and $19$ .
In the original sequence, $a_4=4+3(2.5)=11.5$ and $a_7=4+6(2.5)=19$ , checking the new rules.

Answer

$b_n=4+7.5(n-1)$ for integers $n\ge1$ ; $b_1=4$ , $b_n=b_{n-1}+7.5$ for integers $n\ge2$ .

Key idea

Keeping every third term of an arithmetic sequence triples the difference between successive kept terms.
Problem 4: Mark complete
Problem 5 The matching entry

A sequence is given by $a_n=17-3(n-1)$ for positive integers $n$ . Find every position where the term value equals its own position number.
1. Hint 1
  The position number is also the value required at that position.
2. Hint 2
  Set the term expression equal to $n$ , then check that the resulting position is a positive integer.
Answer

Position $5$ , where $a_5=5$ .

Full solution

The required equality is $a_n=n$ , so
$\begin{aligned}17-3(n-1)&=n\\20-3n&=n\\20&=4n\\n&=5\end{aligned}$

The solution is a positive integer, and
$17-3(5-1)=5$
confirms that the term at position $5$ is $5$ .
The equation has only this solution, so there is no other matching position.

Answer

Position $5$ , where $a_5=5$ .

Key idea

A term that equals its own position is found by equating the term rule with the position variable.
Problem 5: Mark complete
Problem 6 The tile panels

The figure shows the first three panels of a tile pattern. Each later panel repeats the shaded column unchanged and adds one more column of the same size as the columns already beside it, with no overlaps. Write the tile count $a_n$ for Panel $n$ as an explicit rule in the form $a_n=pn+q$ , and find the count in Panel $12$ .

The first three panels of the pattern, with the repeated left column shaded.

Text description of this figure

Three tile panels stand side by side, labeled Panel 1, Panel 2 and Panel 3 from left to right. Every panel begins on the left with the same lightly shaded column of five equal square tiles. To the right of that column, Panel 1 has one unshaded column of three tiles, Panel 2 has two unshaded columns of three tiles, and Panel 3 has three unshaded columns of three tiles. All the columns in a panel rest on the same bottom line, and every tile is the same size, with its boundary drawn. No totals, measurements or rules are printed, and no further panel is shown.
1. Hint 1
  Compare the three panels tile by tile: one part never changes.
2. Hint 2
  Each added column contributes the same count, while the left column contributes once.
Answer

$a_n=3n+5$ for integers $n\ge1$ ; Panel $12$ has $41$ tiles.

Full solution

In every panel the shaded column holds five tiles.
Beside it, Panel $1$ has one unshaded column of three tiles, Panel $2$ has two and Panel $3$ has three, so Panel $n$ has $n$ of those columns.
Therefore
$a_n=3n+5$

The counts are arithmetic: the first count is $8$ and each new panel adds $3$ .
In the requested panel,
$a_{12}=3(12)+5$
so $a_{12}=41$ tiles.

The equivalent arithmetic form is $8+3(n-1)$ .
It gives $8+3(11)=41$ for Panel $12$ , and the first three counts are $8$ , $11$ , and $14$ , matching the figure.

Answer

$a_n=3n+5$ for integers $n\ge1$ ; Panel $12$ has $41$ tiles.

Key idea

A fixed piece plus equally sized added pieces produces an arithmetic sequence of counts.
Problem 6: Mark complete
Problem 7 The reversed strip

A strip has nine labels in order, with values $a_n=-3+2.5(n-1)$ for integers $1\le n\le9$ . The strip is turned end for end, and the labels in the new order are called $b_1$ through $b_9$ . Give $b_n$ in the form $b_n=b_1+(n-1)D$ , and find the new position of the label $9.5$ .
1. Hint 1
  The old last term becomes the new first, and each step now travels backward through the old list.
2. Hint 2
  Find the old ninth value and reverse the sign of the old difference.
3. Hint 3
  Set the new term rule equal to the label value and solve for its position.
Answer

$b_n=17-2.5(n-1)$ for integers $1\le n\le9$ ; $9.5$ is in position $4$ .

Full solution

The old last value is
$a_9=-3+2.5(8)$
so $a_9=17$ .
Moving backward subtracts $2.5$ each time, giving
$b_n=17-2.5(n-1)$

For the requested label,
$\begin{aligned}17-2.5(n-1)&=9.5\\2.5(n-1)&=7.5\\n-1&=3\\n&=4\end{aligned}$

The original sixth label is
$-3+2.5(5)=9.5.$
Reversing nine labels sends position $6$ to position $9-6+1=4$ , checking the result.

Answer

$b_n=17-2.5(n-1)$ for integers $1\le n\le9$ ; $9.5$ is in position $4$ .

Key idea

Reversing a finite arithmetic sequence makes its last value the new start and changes the sign of its difference.
Problem 7: Mark complete
Problem 8 The seven entries

A list has seven real entries $a_1$ through $a_7$ and satisfies $a_3+a_7=2a_5$ . Jo says this guarantees that the whole list is arithmetic. Decide whether Jo is right, and justify your decision.
1. Hint 1
  A condition on a few entries need not control every consecutive difference.
2. Hint 2
  Keep the three entries in the displayed equality compatible, then consider changing an entry absent from that equality.
Answer

No; one example is $1,2,3,4,5,9,7$ .

Full solution

For the example $1,2,3,4,5,9,7$ , the required entries satisfy
$3+7=2(5)$
so the stated condition holds.

But the first difference is $2-1=1$ , while the change from the fifth entry to the sixth is $9-5=4$ .
Since these differences disagree, the list is not arithmetic.
This example disproves the guarantee.

Answer

No; one example is $1,2,3,4,5,9,7$ .

Key idea

An equality involving selected entries does not by itself guarantee a constant difference throughout a list.
Problem 8: Mark complete
Problem 9 The stack of cups

Measured heights of a stack of nested cups are: one cup, $9.5$ centimeters; two cups, $11.1$ centimeters; three cups, $12.7$ centimeters; four cups, $14.3$ centimeters. Each further cup adds the same amount. Write an explicit rule $h_n$ for the height of a stack of $n$ cups, find the height of a stack of $20$ cups, and decide whether any stack is exactly $30$ centimeters tall.
1. Hint 1
  Each extra cup raises the stack by the same amount, so the measured heights are the terms of an arithmetic sequence whose position counts the cups.
2. Hint 2
  Find that constant rise, then build a stack of $n$ cups from the one-cup height and one fewer rises than the number of cups.
3. Hint 3
  A height occurs only when solving the rule for the number of cups gives a whole number.
Answer

$h_n=1.6n+7.9$ centimeters, or equivalently $h_n=9.5+1.6(n-1)$ , for integers $n\ge1$ ; a stack of $20$ cups is $39.9$ centimeters tall; $30$ centimeters would need $n=13.8125$ , which is not a whole number, so no stack has that height.

Full solution

The consecutive gaps are $11.1-9.5=1.6$ , then $12.7-11.1=1.6$ , then
$14.3-12.7=1.6.$
The heights are therefore arithmetic, with first term $9.5$ and common difference $1.6$ .

A stack of $n$ cups takes $n-1$ rises from the one-cup height.
$\begin{aligned}h_n&=9.5+1.6(n-1)\\h_n&=1.6n+7.9\end{aligned}$

For twenty cups, put $n=20$ in that rule.
$h_{20}=1.6(20)+7.9$
So the stack stands $39.9$ centimeters tall.

A stack exactly $30$ centimeters tall would need
$\begin{aligned}1.6n+7.9&=30\\1.6n&=22.1\\n&=13.8125\end{aligned}$

Since $n$ counts cups it has to be a whole number, and $13.8125$ is not one, so that height never occurs.
The nearest stacks hold $13$ cups at $28.7$ centimeters and $14$ cups at $30.3$ centimeters.

Answer

$h_n=1.6n+7.9$ centimeters, or equivalently $h_n=9.5+1.6(n-1)$ , for integers $n\ge1$ ; a stack of $20$ cups is $39.9$ centimeters tall; $30$ centimeters would need $n=13.8125$ , which is not a whole number, so no stack has that height.

Key idea

A constant rise per item makes the readings arithmetic, and a proposed value belongs to them only when its position works out to a whole number.
Problem 9: Mark complete
Problem 10 The constant record

For positive integers $n$ , a sequence has $a_n=8+0(n-1)$ . Setting $a_n=8$ , Ivo reaches $0=0(n-1)$ . He cancels the zero from both sides to get $n-1=0$ , and reports that only $n=1$ works. Is his conclusion correct? State every position with value $8$ and every position with value $9$ .
1. Hint 1
  Evaluate what the rule does at an arbitrary allowed position before judging the step Ivo took.
2. Hint 2
  A zero coefficient leaves the output unchanged, and division by zero is undefined.
Answer

No. Every positive integer position has value $8$ ; no position has value $9$ .

Full solution

For every allowed $n$ , $0(n-1)=0$ , so
$a_n=8$
The equation asking for value $8$ becomes $8=8$ and holds at every positive integer position.

Asking for value $9$ would require $8=9$ , which is false at every position.
Canceling the zero from $0=0(n-1)$ divides both sides by zero, which is undefined, so that step cannot restrict the positions to $n=1$ .

Answer

No. Every positive integer position has value $8$ ; no position has value $9$ .

Key idea

A zero common difference produces a constant sequence, whose membership questions must be settled without dividing by that difference.
Problem 10: Mark complete

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