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Arithmetic Sequences and 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 A rule in squares

The sequence $a_n=(n+2)^2-n^2$ is defined for positive integers $n$ . Find its common difference.
1. Hint 1
  Simplify the rule before comparing consecutive terms.
2. Hint 2
  The quadratic terms cancel when the difference of squares is expanded.
Answer

$d=4$ .

Full solution

Expansion gives
$a_n=4n+4$
Therefore replacing $n$ with $n+1$ increases the value by $4$ for every index.
The first three terms are $8,12,16$ , consistent with that difference.

Answer

$d=4$ .

Key idea

A rule that simplifies to a linear expression in the index defines an arithmetic sequence.
Problem 1: Mark complete
Problem 2 A two-step change

In an arithmetic sequence, $a_{20}-a_7=39$ . Find $a_{101}-a_{99}$ .
1. Hint 1
  The given change covers thirteen equal steps.
2. Hint 2
  The requested change covers two steps of that same size.
Answer

$6$ .

Full solution

The common difference is $d=39/13=3$ , since the index gap is $20-7=13$ .
The requested gap is $101-99=2$ , so
$a_{101}-a_{99}=2(3)$
giving $6$ .

Answer

$6$ .

Key idea

Differences between separated arithmetic terms depend on their index gap rather than their absolute positions.
Problem 2: Mark complete
Problem 3 A length from a total

A finite arithmetic sequence starts at $-7$ , ends at $17$ , and has sum $95$ . How many terms does it contain?
1. Hint 1
  The average of an arithmetic list equals the average of its two endpoints.
2. Hint 2
  Divide the total by that average.
Answer

$19$ terms.

Full solution

The endpoint average is $\frac{-7+17}2=5$ , and the sum of an arithmetic list is its number of terms times that average, so $95=5n$ and $n=19$ .

As a check, $19$ terms from $-7$ to $17$ take $18$ equal steps, so $d=\frac{24}{18}=\frac43$ , a consistent common difference.

Answer

$19$ terms.

Key idea

A nonzero endpoint average determines an arithmetic list's length from its total.
Problem 3: Mark complete
Problem 4 Two conditions on entries

An arithmetic sequence satisfies $a_2+a_8=22$ and $a_8-a_2=-42$ . Find its first term and common difference, and check both conditions.
1. Hint 1
  First recover the two terms from their combined value and their difference.
2. Hint 2
  Adding the two given equations removes one of the unknown terms.
3. Hint 3
  Count the steps between indices $2$ and $8$ before finding the common difference.
Answer

$a_1=39$ and $d=-7$ .

Full solution

Adding the two conditions gives
$\begin{aligned} 2a_8&=-20 \\ a_8&=-10. \end{aligned}$
The sum condition then gives $a_2=22-(-10)$ , so $a_2=32$ .

There are $8-2=6$ steps between the two terms.
Each step has the same difference, so
$\begin{aligned} d&=\frac{-10-32}{6} \\ &=-7. \end{aligned}$
The first term is one step before $a_2$ , giving $a_1=32+7$ , that is, $39$ .

For a check, this sequence gives $a_2=39-7=32$ and $a_8=39+7(-7)$ , which is $-10$ .
Their sum is $22$ and their difference is $-42$ , as required.

Answer

$a_1=39$ and $d=-7$ .

Key idea

Combined information about two terms can be separated before their index gap determines the common difference.
Problem 4: Mark complete
Problem 5 A descending schedule

A printer schedules batches of $46$ labels at first and $3$ fewer labels in each following batch, ending with a batch of $16$ labels. Planned batch $4$ is canceled, and all the remaining planned batches are printed at their planned sizes. How many batches are printed, and how many labels are printed altogether?
1. Hint 1
  Work out the complete planned arithmetic schedule first.
2. Hint 2
  Then remove the canceled batch from both the count and the total.
Answer

$10$ batches; $304$ labels.

Full solution

The planned difference is $d=-3$ .
The number of steps is $(16-46)/(-3)=10$ , so the plan has $11$ batches, and its total is
$S=\frac{11}{2}(46+16)$
which is $341$ labels.

Planned batch $4$ holds $46+3(-3)=37$ labels.
Removing it leaves $10$ batches and $341-37=304$ labels.

Answer

$10$ batches; $304$ labels.

Key idea

Counting arithmetic terms requires one more than the number of equal changes between the endpoints, before any term is removed.
Problem 5: Mark complete
Problem 6 Production and a rising target

The total produced after $n$ days is $S_n=3n^2+8n$ items for integers $n\ge0$ . The daily target is $48$ items on day one and rises by $4$ items each day. Find the common difference in daily production and the first day on which production meets or exceeds that day's target. State the production and the target on that day.
1. Hint 1
  The amount made on a day is the increase in the cumulative total on that day.
2. Hint 2
  Subtract successive totals, write the daily target as an arithmetic term rule, and solve the comparison for a whole-number day.
Answer

$d=6$ items; day $20$ , with production $125$ items and target $124$ items.

Full solution

The daily production is
$a_n=S_n-S_{n-1}.$
Since $S_{n-1}=3n^2+2n-5$ , this gives $a_n=6n+5$ for $n\ge1$ , which rises by $d=6$ items per day.

The target on day $n$ is
$48+4(n-1)=4n+44.$
Production meets it when $6n+5\ge4n+44$ , that is, $2n\ge39$ , so the first such day is $n=20$ .

On day $20$ production is $125$ items against a target of $124$ .
On day $19$ it was $119$ against $120$ , so day $20$ is the first.

Answer

$d=6$ items; day $20$ , with production $125$ items and target $124$ items.

Key idea

Successive differences of a quadratic cumulative total give a linear term rule that can be compared with another arithmetic rule.
Problem 6: Mark complete
Problem 7 An incomplete total

A record pairs the $12$ terms of an arithmetic sequence from the ends inward. It includes $a_1+a_{12}$ , $a_2+a_{11}$ , $a_3+a_{10}$ , $a_4+a_9$ , and $a_5+a_8$ , but omits $a_6+a_7$ . The five recorded pair totals add to $170$ . Find the missing pair total and the sum $S_{12}$ of all $12$ terms. Justify why every pair has the same total, and explain how adding the sequence in forward and reverse order gives the same sum.
1. Hint 1
  Moving one position inward increases one member of a pair by the common difference and decreases the other by that amount.
2. Hint 2
  Express $a_k+a_{13-k}$ in terms of the first term and common difference.
3. Hint 3
  Use the five recorded pairs to recover the common pair total; forward and reversed rows together count the sum twice.
Answer

Missing pair total $34$ ; $S_{12}=204$ .

Full solution

Let the common difference be $d$ .
For an end-to-end pair, the two terms are $a_1+(k-1)d$ and $a_1+(12-k)d$ .
Their step counts add to $11$ , so
$a_k+a_{13-k}=2a_1+11d.$
The right side does not depend on $k$ , proving that every pair has the same total.
Call that total $C$ .

The five recorded totals give
$\begin{aligned} 5C&=170 \\ C&=34. \end{aligned}$
Thus the omitted pair also totals $34$ .
The six pairs cover every term once, so their complete sum is $6(34)=204$ .

Writing all $12$ terms forward and then in reverse produces $12$ columns, each totaling $C$ , and counts the original sum twice.
Therefore
$\begin{aligned} 2S_{12}&=12(34) \\ S_{12}&=204. \end{aligned}$
This agrees with restoring the missing pair:
$170+34=204.$

Answer

Missing pair total $34$ ; $S_{12}=204$ .

Key idea

Equal end-to-end pair totals let an incomplete arithmetic-sum record determine the whole sum.
Problem 7: Mark complete
Problem 8 An average claim

For an arithmetic sequence with $15$ terms, a student claims its average equals $a_8$ . Is this true even if some terms are negative? Explain.
1. Hint 1
  Compare terms the same distance from the middle index.
2. Hint 2
  Each such pair differs from $a_8$ by opposite multiples of $d$ , which may both be zero.
Answer

Yes; the average is $a_8$ .

Full solution

The pairs $a_{8-j}$ and $a_{8+j}$ total $2a_8$ for
$j=1,2,\ldots,7.$
Seven pairs plus the middle term give
$S_{15}=15a_8$
Dividing by $15$ gives average $a_8$ .
This argument uses differences and works regardless of the terms' signs.

Answer

Yes; the average is $a_8$ .

Key idea

The average of an odd-length arithmetic list is its middle term, including when terms are negative.
Problem 8: Mark complete
Problem 9 A total with a constant

A sequence has $S_0=0$ and $S_n=n^2+2n+5$ for every positive integer $n$ . A student says the sequence must be arithmetic because this is a quadratic formula for its positive-index totals. Is that correct? Justify your answer with the first three terms.
1. Hint 1
  The first term is $S_1-S_0$ , so use the separately stated value of $S_0$ .
2. Hint 2
  Compare the differences between the three terms you recover.
Answer

No; the first three terms are $8,5,7$ , with differences $-3$ and $2$ .

Full solution

The totals are $S_1=8$ , $S_2=13$ , and $S_3=20$ .
Therefore the first three terms are $8$ , $13-8=5$ , and $20-13=7$ .

Their differences are $-3$ and $2$ , which disagree.
The constant $5$ makes the positive-index formula inconsistent with the zero total at index zero; the arithmetic-total form has no constant term.

Answer

No; the first three terms are $8,5,7$ , with differences $-3$ and $2$ .

Key idea

A quadratic formula for positive-index totals needs the correct zero total before it guarantees an arithmetic sequence.
Problem 9: Mark complete
Problem 10 A shifted list

An arithmetic sequence has common difference $-4$ . A student adds $10$ to every term and says the new common difference is $6$ . Is that correct? Give the new common difference and explain.
1. Hint 1
  Find a neighboring difference after both terms receive the same added constant.
2. Hint 2
  The two added constants cancel when the new terms are subtracted.
Answer

No; the new common difference is $-4$ .

Full solution

Write
$b_n=a_n+10.$
Then
$b_{n+1}-b_n=a_{n+1}-a_n$
because the added $10$ cancels from the subtraction.
The difference remains $-4$ ; adding $10$ moves the entire list without changing the gaps.

Answer

No; the new common difference is $-4$ .

Key idea

Adding the same constant to every arithmetic term preserves its common difference.
Problem 10: Mark complete

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