Practice · Find the rule in a fraction sequence · Sensim Math

Variant 1 easy

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 5th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 5th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 5th fraction.

Unknowns

The fraction at position 5.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 5 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6.

1,\ 3,\ 6

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 5

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 3 the total is 3 fractions (positions 1 through 3). The next group is denominator 4 (3 fractions, positions 4 through 6). So position 5 is in the denominator-4 group.

3 < 5 \le 6 \Rightarrow \text{denominator } 4

Using the running totals reduces a far-off 5th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 5 is the (5 - 3) = 2th fraction in the denominator-4 group, whose numerators count 1, 2, 3, ... So the numerator is 2.

5 - 3 = 2 \Rightarrow \frac{2}{4}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{2}{4}

4 · Reviewdoes it hold up?

Positions 4-6 hold $\frac{1}{4}$ through $\frac{3}{4}$ ; the 5th is the 2th of these, which is $\frac{2}{4}$ . This fits the pattern (numerator counts up, denominator fixed at 4). The numerator 2 is between 1 and 3, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-4 group $\frac{1}{4}$ , $\frac{2}{4}$ , ... from position 4; the 2th lands on position 5, confirming $\frac{2}{4}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 5.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 5, then count inside that group!

Variant 2 easy

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 7th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 7th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 7th fraction.

Unknowns

The fraction at position 7.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 7 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10.

1,\ 3,\ 6,\ 10

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 7

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 4 the total is 6 fractions (positions 1 through 6). The next group is denominator 5 (4 fractions, positions 7 through 10). So position 7 is in the denominator-5 group.

6 < 7 \le 10 \Rightarrow \text{denominator } 5

Using the running totals reduces a far-off 7th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 7 is the (7 - 6) = 1th fraction in the denominator-5 group, whose numerators count 1, 2, 3, ... So the numerator is 1.

7 - 6 = 1 \Rightarrow \frac{1}{5}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{1}{5}

4 · Reviewdoes it hold up?

Positions 7-10 hold $\frac{1}{5}$ through $\frac{4}{5}$ ; the 7th is the 1th of these, which is $\frac{1}{5}$ . This fits the pattern (numerator counts up, denominator fixed at 5). The numerator 1 is between 1 and 4, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-5 group $\frac{1}{5}$ , $\frac{2}{5}$ , ... from position 7; the 1th lands on position 7, confirming $\frac{1}{5}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 7.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 7, then count inside that group!

Variant 3 easy

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 8th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 8th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 8th fraction.

Unknowns

The fraction at position 8.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 8 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10.

1,\ 3,\ 6,\ 10

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 8

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 4 the total is 6 fractions (positions 1 through 6). The next group is denominator 5 (4 fractions, positions 7 through 10). So position 8 is in the denominator-5 group.

6 < 8 \le 10 \Rightarrow \text{denominator } 5

Using the running totals reduces a far-off 8th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 8 is the (8 - 6) = 2th fraction in the denominator-5 group, whose numerators count 1, 2, 3, ... So the numerator is 2.

8 - 6 = 2 \Rightarrow \frac{2}{5}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{2}{5}

4 · Reviewdoes it hold up?

Positions 7-10 hold $\frac{1}{5}$ through $\frac{4}{5}$ ; the 8th is the 2th of these, which is $\frac{2}{5}$ . This fits the pattern (numerator counts up, denominator fixed at 5). The numerator 2 is between 1 and 4, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-5 group $\frac{1}{5}$ , $\frac{2}{5}$ , ... from position 7; the 2th lands on position 8, confirming $\frac{2}{5}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 8.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 8, then count inside that group!

Variant 4 easy

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 13th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 13th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 13th fraction.

Unknowns

The fraction at position 13.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 13 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15.

1,\ 3,\ 6,\ 10,\ 15

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 13

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 5 the total is 10 fractions (positions 1 through 10). The next group is denominator 6 (5 fractions, positions 11 through 15). So position 13 is in the denominator-6 group.

10 < 13 \le 15 \Rightarrow \text{denominator } 6

Using the running totals reduces a far-off 13th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 13 is the (13 - 10) = 3th fraction in the denominator-6 group, whose numerators count 1, 2, 3, ... So the numerator is 3.

13 - 10 = 3 \Rightarrow \frac{3}{6}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{3}{6}

4 · Reviewdoes it hold up?

Positions 11-15 hold $\frac{1}{6}$ through $\frac{5}{6}$ ; the 13th is the 3th of these, which is $\frac{3}{6}$ . This fits the pattern (numerator counts up, denominator fixed at 6). The numerator 3 is between 1 and 5, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-6 group $\frac{1}{6}$ , $\frac{2}{6}$ , ... from position 11; the 3th lands on position 13, confirming $\frac{3}{6}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 13.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 13, then count inside that group!

Variant 5 medium

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 15th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 15th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 15th fraction.

Unknowns

The fraction at position 15.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 15 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15.

1,\ 3,\ 6,\ 10,\ 15

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 15

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 5 the total is 10 fractions (positions 1 through 10). The next group is denominator 6 (5 fractions, positions 11 through 15). So position 15 is in the denominator-6 group.

10 < 15 \le 15 \Rightarrow \text{denominator } 6

Using the running totals reduces a far-off 15th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 15 is the (15 - 10) = 5th fraction in the denominator-6 group, whose numerators count 1, 2, 3, ... So the numerator is 5.

15 - 10 = 5 \Rightarrow \frac{5}{6}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{5}{6}

4 · Reviewdoes it hold up?

Positions 11-15 hold $\frac{1}{6}$ through $\frac{5}{6}$ ; the 15th is the 5th of these, which is $\frac{5}{6}$ . This fits the pattern (numerator counts up, denominator fixed at 6). The numerator 5 is between 1 and 5, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-6 group $\frac{1}{6}$ , $\frac{2}{6}$ , ... from position 11; the 5th lands on position 15, confirming $\frac{5}{6}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 15.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 15, then count inside that group!

Variant 6 medium

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 20th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 20th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 20th fraction.

Unknowns

The fraction at position 20.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 20 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15, 21.

1,\ 3,\ 6,\ 10,\ 15,\ 21

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 20

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 6 the total is 15 fractions (positions 1 through 15). The next group is denominator 7 (6 fractions, positions 16 through 21). So position 20 is in the denominator-7 group.

15 < 20 \le 21 \Rightarrow \text{denominator } 7

Using the running totals reduces a far-off 20th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 20 is the (20 - 15) = 5th fraction in the denominator-7 group, whose numerators count 1, 2, 3, ... So the numerator is 5.

20 - 15 = 5 \Rightarrow \frac{5}{7}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{5}{7}

4 · Reviewdoes it hold up?

Positions 16-21 hold $\frac{1}{7}$ through $\frac{6}{7}$ ; the 20th is the 5th of these, which is $\frac{5}{7}$ . This fits the pattern (numerator counts up, denominator fixed at 7). The numerator 5 is between 1 and 6, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-7 group $\frac{1}{7}$ , $\frac{2}{7}$ , ... from position 16; the 5th lands on position 20, confirming $\frac{5}{7}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 20.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 20, then count inside that group!

Variant 7 medium

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 28th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 28th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 28th fraction.

Unknowns

The fraction at position 28.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 28 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15, 21, 28.

1,\ 3,\ 6,\ 10,\ 15,\ 21,\ 28

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 28

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 7 the total is 21 fractions (positions 1 through 21). The next group is denominator 8 (7 fractions, positions 22 through 28). So position 28 is in the denominator-8 group.

21 < 28 \le 28 \Rightarrow \text{denominator } 8

Using the running totals reduces a far-off 28th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 28 is the (28 - 21) = 7th fraction in the denominator-8 group, whose numerators count 1, 2, 3, ... So the numerator is 7.

28 - 21 = 7 \Rightarrow \frac{7}{8}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{7}{8}

4 · Reviewdoes it hold up?

Positions 22-28 hold $\frac{1}{8}$ through $\frac{7}{8}$ ; the 28th is the 7th of these, which is $\frac{7}{8}$ . This fits the pattern (numerator counts up, denominator fixed at 8). The numerator 7 is between 1 and 7, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-8 group $\frac{1}{8}$ , $\frac{2}{8}$ , ... from position 22; the 7th lands on position 28, confirming $\frac{7}{8}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 28.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 28, then count inside that group!

Variant 8 medium

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 36th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 36th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 36th fraction.

Unknowns

The fraction at position 36.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 36 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15, 21, 28, 36.

1,\ 3,\ 6,\ 10,\ 15,\ 21,\ 28,\ 36

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 36

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 8 the total is 28 fractions (positions 1 through 28). The next group is denominator 9 (8 fractions, positions 29 through 36). So position 36 is in the denominator-9 group.

28 < 36 \le 36 \Rightarrow \text{denominator } 9

Using the running totals reduces a far-off 36th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 36 is the (36 - 28) = 8th fraction in the denominator-9 group, whose numerators count 1, 2, 3, ... So the numerator is 8.

36 - 28 = 8 \Rightarrow \frac{8}{9}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{8}{9}

4 · Reviewdoes it hold up?

Positions 29-36 hold $\frac{1}{9}$ through $\frac{8}{9}$ ; the 36th is the 8th of these, which is $\frac{8}{9}$ . This fits the pattern (numerator counts up, denominator fixed at 9). The numerator 8 is between 1 and 8, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-9 group $\frac{1}{9}$ , $\frac{2}{9}$ , ... from position 29; the 8th lands on position 36, confirming $\frac{8}{9}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 36.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 36, then count inside that group!

Variant 9 hard

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 41st position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 41st position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 41st fraction.

Unknowns

The fraction at position 41.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 41 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15, 21, 28, 36, 45.

1,\ 3,\ 6,\ 10,\ 15,\ 21,\ 28,\ 36,\ 45

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 41

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 9 the total is 36 fractions (positions 1 through 36). The next group is denominator 10 (9 fractions, positions 37 through 45). So position 41 is in the denominator-10 group.

36 < 41 \le 45 \Rightarrow \text{denominator } 10

Using the running totals reduces a far-off 41st term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 41 is the (41 - 36) = 5th fraction in the denominator-10 group, whose numerators count 1, 2, 3, ... So the numerator is 5.

41 - 36 = 5 \Rightarrow \frac{5}{10}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{5}{10}

4 · Reviewdoes it hold up?

Positions 37-45 hold $\frac{1}{10}$ through $\frac{9}{10}$ ; the 41st is the 5th of these, which is $\frac{5}{10}$ . This fits the pattern (numerator counts up, denominator fixed at 10). The numerator 5 is between 1 and 9, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-10 group $\frac{1}{10}$ , $\frac{2}{10}$ , ... from position 37; the 5th lands on position 41, confirming $\frac{5}{10}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 41.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 41, then count inside that group!

Variant 10 hard

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 50th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 50th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 50th fraction.

Unknowns

The fraction at position 50.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 50 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55.

1,\ 3,\ 6,\ 10,\ 15,\ 21,\ 28,\ 36,\ 45,\ 55

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 50

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 10 the total is 45 fractions (positions 1 through 45). The next group is denominator 11 (10 fractions, positions 46 through 55). So position 50 is in the denominator-11 group.

45 < 50 \le 55 \Rightarrow \text{denominator } 11

Using the running totals reduces a far-off 50th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 50 is the (50 - 45) = 5th fraction in the denominator-11 group, whose numerators count 1, 2, 3, ... So the numerator is 5.

50 - 45 = 5 \Rightarrow \frac{5}{11}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{5}{11}

4 · Reviewdoes it hold up?

Positions 46-55 hold $\frac{1}{11}$ through $\frac{10}{11}$ ; the 50th is the 5th of these, which is $\frac{5}{11}$ . This fits the pattern (numerator counts up, denominator fixed at 11). The numerator 5 is between 1 and 10, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-11 group $\frac{1}{11}$ , $\frac{2}{11}$ , ... from position 46; the 5th lands on position 50, confirming $\frac{5}{11}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 50.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 50, then count inside that group!

Variant 11 hard

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 66th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 66th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 66th fraction.

Unknowns

The fraction at position 66.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 66 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66.

1,\ 3,\ 6,\ 10,\ 15,\ 21,\ 28,\ 36,\ 45,\ 55,\ 66

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 66

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 11 the total is 55 fractions (positions 1 through 55). The next group is denominator 12 (11 fractions, positions 56 through 66). So position 66 is in the denominator-12 group.

55 < 66 \le 66 \Rightarrow \text{denominator } 12

Using the running totals reduces a far-off 66th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 66 is the (66 - 55) = 11th fraction in the denominator-12 group, whose numerators count 1, 2, 3, ... So the numerator is 11.

66 - 55 = 11 \Rightarrow \frac{11}{12}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{11}{12}

4 · Reviewdoes it hold up?

Positions 56-66 hold $\frac{1}{12}$ through $\frac{11}{12}$ ; the 66th is the 11th of these, which is $\frac{11}{12}$ . This fits the pattern (numerator counts up, denominator fixed at 12). The numerator 11 is between 1 and 11, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-12 group $\frac{1}{12}$ , $\frac{2}{12}$ , ... from position 56; the 11th lands on position 66, confirming $\frac{11}{12}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 66.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 66, then count inside that group!

Variant 12 hard

The fractions below are listed following a fixed rule. Find the fraction that will be placed in the 100th position.

$\frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ \frac{1}{4},\ \frac{2}{4},\ \frac{3}{4},\ \frac{1}{5},\ \cdots$

Your answer

Show solution

1 · Understandwhat's really being asked

Fractions are listed by a rule: first all fractions with denominator 2, then denominator 3, then 4, and so on, with numerators counting up from 1 to one less than the denominator within each group. I need the fraction in the 100th position.

Givens

The list is $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{4}$ , $\frac{2}{4}$ , $\frac{3}{4}$ , $\frac{1}{5}$ , ...
Group with denominator (k+1) has k fractions, with numerators 1, 2, ..., k.
I want the 100th fraction.

Unknowns

The fraction at position 100.

Constraints

Within a group the denominator is fixed and numerators run 1 up to denominator minus 1.
Groups appear in order of increasing denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The list groups by denominator, and the group with denominator (k+1) contributes k terms. Adding up these group sizes (1+2+3+...) tells me which group position 100 falls in, then I count within that group.

3 · Execute3 carry out the plan

1See the group sizes

#5 Look for a Pattern 3.OA.D.9

Denominator 2 has 1 fraction, denominator 3 has 2, denominator 4 has 3, and in general denominator d has d-1 fractions. The running totals are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105.

1,\ 3,\ 6,\ 10,\ 15,\ 21,\ 28,\ 36,\ 45,\ 55,\ 66,\ 78,\ 91,\ 105

Each new denominator adds one more fraction than the previous group, a clear growing pattern.

2Locate group for position 100

#9 Solve an Easier Related Problem 3.OA.D.9

After denominator 14 the total is 91 fractions (positions 1 through 91). The next group is denominator 15 (14 fractions, positions 92 through 105). So position 100 is in the denominator-15 group.

91 < 100 \le 105 \Rightarrow \text{denominator } 15

Using the running totals reduces a far-off 100th term to 'which short group is it in'.

3Find the position within the group

#5 Look for a Pattern 3.NF.A.1

Position 100 is the (100 - 91) = 9th fraction in the denominator-15 group, whose numerators count 1, 2, 3, ... So the numerator is 9.

100 - 91 = 9 \Rightarrow \frac{9}{15}

Within a group the numerator equals the place number, so the term has that numerator.

Answer:

\frac{9}{15}

4 · Reviewdoes it hold up?

Positions 92-105 hold $\frac{1}{15}$ through $\frac{14}{15}$ ; the 100th is the 9th of these, which is $\frac{9}{15}$ . This fits the pattern (numerator counts up, denominator fixed at 15). The numerator 9 is between 1 and 14, as required for that group.

Another way: Make a systematic list (tool 2): keep listing the denominator-15 group $\frac{1}{15}$ , $\frac{2}{15}$ , ... from position 92; the 9th lands on position 100, confirming $\frac{9}{15}$ .

Standardsmin grade 3

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Finding that group sizes grow by 1 and using running totals to locate position 100.
3.NF.A.1 Understand a fraction as quantity formed by parts of a whole — Reading off the numerator and denominator of the located fraction.

💡Takeaway. This only needs Grade 3 pattern sense: add up the group sizes until you reach 100, then count inside that group!

Find the rule in a fraction sequence · 12 practice problems

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