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Arithmetic 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 new total

The sum of the first $7$ terms of an arithmetic sequence is $46$ . The sum of its first $8$ terms is $55$ . Find its eighth term.
1. Hint 1
  Compare the terms included in the two totals.
2. Hint 2
  Subtract the shorter total from the longer total.
Answer

$a_8=9$ .

Full solution

The longer sum contains exactly one extra term, $a_8$ .
Therefore
$a_8=55-46.$
This gives
$a_8=9.$
The check is $46+9=55$ .

Answer

$a_8=9$ .

Key idea

The difference between consecutive partial sums is the newly included term.
Problem 1: Mark complete
Problem 2 The late start

Evaluate $\sum_{j=6}^{20}(43-3j)$ .
1. Hint 1
  Each step of the index changes the summand by the same amount, so the terms are evenly spaced.
2. Hint 2
  Both limits are included, so count the index values from the lower limit through the upper limit.
3. Hint 3
  Substitute each limit to get the first and last terms, then use the arithmetic sum formula.
Answer

$60$ .

Full solution

Raising the index by $1$ lowers the summand by $3$ , so the terms are evenly spaced.
Substituting the lower limit gives
$43-3\cdot6=25.$
Substituting the upper limit gives
$43-3\cdot20=-17.$
Both limits are included, so the number of terms is $20-6+1$ , which is $15$ .
The first and last terms add to $8$ , so
$S=\frac{15\cdot8}{2}.$
Therefore $S=60$ .

Answer

$60$ .

Key idea

A sigma whose summand changes by a fixed amount per index step is an arithmetic series, so its own first term, last term and count give the total.
Problem 2: Mark complete
Problem 3 The starting value

The first nine terms of an arithmetic sequence total $117$ , and its common difference is $2$ . Find the first term.
1. Hint 1
  The ninth term is eight steps after the first.
2. Hint 2
  Use the arithmetic sum formula with $a_9=a_1+16$ .
Answer

$a_1=5$ .

Full solution

There are eight increases of $2$ , so
$a_9=a_1+16.$
The sum formula gives
$117=\frac{9(2a_1+16)}{2}.$
Dividing by $9$ and multiplying by $2$ gives
$26=2a_1+16.$
Therefore
$a_1=5.$
The last term is $21$ , and
$9(5+21)/2=117.$

Answer

$a_1=5$ .

Key idea

A known sum and common difference can determine the first term of an arithmetic sequence.
Problem 3: Mark complete
Problem 4 The omitted entries

A list has $8$ arithmetic terms, first term $5$ , and common difference $3$ . A report omits the third and sixth entries and includes each of the remaining six entries once. Find the report total.
1. Hint 1
  Find the total of the complete list.
2. Hint 2
  Find the values at the two omitted positions.
3. Hint 3
  Subtract both omitted values from the complete total.
Answer

$93$ .

Full solution

The last entry is
$5+7\cdot3=26.$
The complete total is
$S_8=\frac{8(5+26)}{2}.$
Thus $S_8=124$ .
The omitted entries are
$a_3=5+2\cdot3$
and
$a_6=5+5\cdot3.$
They equal $11$ and $20$ .
The report total is
$124-11-20=93.$

Answer

$93$ .

Key idea

Find a complete arithmetic total first, then remove any entries that were excluded.
Problem 4: Mark complete
Problem 5 The overlapping reports

A record has entries $a_k=2k+1$ . One report totals positions $3$ through $11$ , inclusive; another totals positions $8$ through $15$ , inclusive. Find the amount counted twice when the reports are added, and the total for positions $3$ through $15$ counted once each.
1. Hint 1
  Identify the positions common to both reports.
2. Hint 2
  The overlap runs from the later of the two starting positions to the earlier of the two ending positions.
3. Hint 3
  For the complete range, count both endpoint positions.
Answer

Counted twice: $80$ ; total counted once: $247$ .

Full solution

The overlap consists of positions $8$ through $11$ , whose values run from $17$ to $23$ .
Its total is
$\frac{4(17+23)}{2}=80.$
The union consists of $15-3+1=13$ positions, from value $7$ to value $31$ .
Its total is
$\frac{13(7+31)}{2}=247.$
Adding the two report totals would give $247+80=327$ , because the overlap is included twice.

Answer

Counted twice: $80$ ; total counted once: $247$ .

Key idea

Overlapping index ranges require counting the shared entries only once in the final total.
Problem 5: Mark complete
Problem 6 The distances from zero

Eight points on a number line have coordinates forming an arithmetic sequence with first term $-5$ and common difference $2$ . Find the sum of their distances from zero.
1. Hint 1
  A distance from zero is the absolute value of a coordinate.
2. Hint 2
  Separate the negative coordinates from the positive coordinates.
3. Hint 3
  Add each group of positive distances.
Answer

$34$ units.

Full solution

The coordinates are $-5,-3,-1,1,3,5,7,9$ .
Their distances form the groups $5,3,1$ and $1,3,5,7,9$ .
The first group has sum
$\frac{3(5+1)}{2}=9.$
The second group has sum
$\frac{5(1+9)}{2}=25.$
The total distance is
$9+25=34.$
Adding the signed coordinates instead would measure a different quantity.

Answer

$34$ units.

Key idea

When summing distances, convert signed coordinates to nonnegative distances before adding.
Problem 6: Mark complete
Problem 7 The two totals

An arithmetic sequence has $S_3=18$ and $S_5=45$ , where $S_n$ is the sum of its first $n$ terms. Find its first term, common difference, and $S_7$ .
1. Hint 1
  Write both totals in terms of $a_1$ and $d$ .
2. Hint 2
  Simplify each equation before solving the pair.
3. Hint 3
  Use the recovered first term and difference in the seven-term sum.
Answer

$a_1=3$ , $d=3$ , and $S_7=84$ .

Full solution

Substituting into $S_n=\frac{n}{2}\big(2a_1+(n-1)d\big)$ gives
$18=\frac{3}{2}(2a_1+2d)$
and
$45=\frac{5}{2}(2a_1+4d).$
Dividing the first by $3$ and the second by $5$ gives $a_1+d=6$ and $a_1+2d=9$ .
Subtracting these equations gives
$d=3.$
Then $a_1=3$ .
The seventh term is $3+6\cdot3=21$ , so
$S_7=\frac{7(3+21)}{2}.$
Therefore
$S_7=84.$
The first three terms total $3+6+9=18$ , and the first five total $45$ .

Answer

$a_1=3$ , $d=3$ , and $S_7=84$ .

Key idea

Two partial sums of an arithmetic sequence can determine its first term and common difference.
Problem 7: Mark complete
Problem 8 The shifted second row

An arithmetic row $a_1,\ldots,a_n$ has sum $S$ . A second row reverses it and adds $c$ to every entry. A student says that adding the rows proves $2S=n(a_1+a_n+c)$ . Is this correct for every $c$ ? Use the two rows to derive the correct formula for $S$ .
1. Hint 1
  Compare the sum of the second row with $S$ .
2. Hint 2
  Each column still has the same total.
3. Hint 3
  Account for the added $c$ in all $n$ entries before solving for $S$ .
Answer

No (the equality holds only when $c=0$ ); $S=\frac{n(a_1+a_n)}{2}$ .

Full solution

Reversal does not change a sum, but adding $c$ to every entry makes the second row total $S+nc$ .
Equal distances from the ends of an arithmetic list give
$a_j+a_{n+1-j}=a_1+a_n.$
Set $E=a_1+a_n$ .
Every column therefore totals $E+c$ , and adding all columns gives
$2S+nc=n(E+c).$
Expanding and canceling $nc$ gives
$2S=nE.$
Substituting for $E$ yields
$S=\frac{n(a_1+a_n)}{2}.$
The proposed equality holds when $c=0$ , but omits the change in the second row when $c\ne0$ .

Answer

No (the equality holds only when $c=0$ ); $S=\frac{n(a_1+a_n)}{2}$ .

Key idea

Reversing and adding an arithmetic list pairs equal endpoint sums, but any adjustment to the copied row must also change its total.
Problem 8: Mark complete
Problem 9 The shaded square

The figure shows a $4$ by $4$ array with $1,2,3,4$ cells shaded in successive rows. In an $n$ by $n$ version, row $k$ has its first $k$ cells shaded. Use reflection across the main diagonal to derive a formula for $1+2+\cdots+n$ .

A $4$ by $4$ array whose row $k$ has its first $k$ cells shaded, with the main diagonal outlined.

Text description of this figure

A square array of four rows and four columns of equal cells, with the rows numbered 1 to 4 from top to bottom. In row 1 the first cell from the left is shaded. In row 2 the first two cells are shaded, in row 3 the first three, and in row 4 all four cells are shaded. The four cells running from the top left corner to the bottom right corner, the main diagonal, carry a heavier outline. Every shaded cell uses the same light shade, and no count, total or formula is printed.
1. Hint 1
  The $n$ diagonal cells are all shaded.
2. Hint 2
  Reflection pairs each cell below the diagonal with one above it.
3. Hint 3
  Count half the cells outside the diagonal, then include the diagonal cells.
Answer

$1+2+\cdots+n=\frac{n(n+1)}{2}$ .

Full solution

There are $n^2$ cells altogether and $n$ on the main diagonal.
The cell in row $i$ and column $j$ is shaded exactly when $j\le i$ , so of any off-diagonal cell and its mirror image, exactly one is shaded.
Reflection pairs the other $n^2-n$ cells, with one shaded cell in each pair.
If $T$ is the number of shaded cells, then
$T=\frac{n^2-n}{2}+n.$
Combining the terms gives
$T=\frac{n^2+n}{2}.$
Factoring gives
$T=\frac{n(n+1)}{2}.$
The rows contain $1,2,\ldots,n$ shaded cells, so this is their sum.

Answer

$1+2+\cdots+n=\frac{n(n+1)}{2}$ .

Key idea

Counting the same shaded region by rows and by diagonal symmetry proves the formula for the first $n$ positive integers.
Problem 9: Mark complete
Problem 10 The zero total

A list has an odd positive number of arithmetic terms and total $0$ . Theo says its middle term must be $0$ . Is he correct? Justify your answer, and say whether your argument depends on the sign of the common difference.
1. Hint 1
  Write the term count as $2m+1$ , where $m\ge0$ .
2. Hint 2
  Terms equally far from the middle have average equal to the middle term.
3. Hint 3
  Express the whole sum using the middle term and the term count.
Answer

Yes; the middle term is $0$ , and the argument does not depend on the sign of the common difference.

Full solution

Let the middle term be $b$ and the count be $2m+1$ .
Terms $j$ positions from the middle are $b-jd$ and $b+jd$ , whose sum is $2b$ .
There are $m$ such pairs and the middle term itself, so
$S=2mb+b.$
Since $S=0$ ,
$0=(2m+1)b.$
The count $2m+1$ is positive, so $b=0$ .
The argument also covers a one-term list and does not depend on the sign of $d$ .

Answer

Yes; the middle term is $0$ , and the argument does not depend on the sign of the common difference.

Key idea

The average of an odd-length arithmetic list equals its middle term.
Problem 10: Mark complete

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