Math worksheet for identifying growing patterns in sequences.

Let's solve each pattern step by step and determine what comes next in each sequence.

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1. 10, 20, 30, 40, 50, 60, 70, 80, ___

- This is a simple arithmetic sequence where each number increases by 10.
- 80 + 10 = 90

✔ Answer: 90

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2. A, B, A, A, B, B, A, A, A, B, B, B, ___

- Let's look at the pattern:
- A, B
- A, A, B, B
- A, A, A, B, B, B

It appears that:
- The number of A's increases by one each time (1, 2, 3)
- The number of B's also increases by one each time (1, 2, 3)

So the next group should be:
- Four A's followed by four B's → A, A, A, A, B, B, B, B

But we are only looking for the next single letter after "B, B, B".

The last three letters are: A, A, A, B, B, B
So the next should be A, starting the next block of four A’s?

Wait — let's list the full sequence with positions:

| Position | Letter |
|---------|--------|
| 1 | A |
| 2 | B |
| 3 | A |
| 4 | A |
| 5 | B |
| 6 | B |
| 7 | A |
| 8 | A |
| 9 | A |
| 10 | B |
| 11 | B |
| 12 | B |

Now, we can see:
- First group: A, B (length 2)
- Second group: A,A,B,B (length 4)
- Third group: A,A,A,B,B,B (length 6)

Each group has increasing numbers of A and B, with equal counts.

So the next group should be:
- Four A’s and four B’s → A,A,A,A,B,B,B,B

After the third group ends at position 12 with B, the next (13th) letter should be the first A of the next group.

✔ Answer: A

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3. 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, ___

- These are odd numbers, increasing by 2.
- 19 + 2 = 21

✔ Answer: 21

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4. A, B, A, A, B, A, A, A, B, A, A, A, A, B, ___

Let’s break this down:

Look at the positions of B:
- B at position 2
- B at position 5
- B at position 8
- B at position 11
- B at position 14

So B appears every 3rd position starting from 2: 2, 5, 8, 11, 14 → next is 17

So the next letter (position 15) is A, then position 16: A, position 17: B

But the question is: what comes next after “A, A, A, A, B”?

Let’s write the sequence:

| Pos | Letter |
|-----|--------|
| 1 | A |
| 2 | B |
| 3 | A |
| 4 | A |
| 5 | B |
| 6 | A |
| 7 | A |
| 8 | A |
| 9 | B |
|10 | A |
|11 | A |
|12 | A |
|13 | A |
|14 | B |
|15 | ? |

We see that B appears at positions: 2, 5, 8, 11, 14 → every 3 steps

So next B is at 17, so positions 15 and 16 are A’s.

Thus, after "A, A, A, A, B" (positions 13–14), the next letter (position 15) is A

✔ Answer: A

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5. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ___

- Even numbers increasing by 2.
- 20 + 2 = 22

✔ Answer: 22

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6. 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, ___

Look at the pattern:
- 1 → once
- 2 → twice
- 3 → twice
- 4 → three times
- 5 → four times?

Wait, let's check:

| Number | Count |
|--------|-------|
| 1 | 1 |
| 2 | 2 |
| 3 | 2 |
| 4 | 3 |
| 5 | 4? |

Wait — but the sequence is:

1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5

So:
- 1 → 1 time
- 2 → 2 times
- 3 → 2 times
- 4 → 3 times
- 5 → 4 times

Hmm — not consistent.

Wait — perhaps it's based on how many times each number repeats?

Let’s count:

- 1: once → 1
- 2: twice → 2
- 3: twice → 2
- 4: three times → 3
- 5: four times → 4

But why does 3 repeat twice? Maybe it's not about the number, but about the number of repetitions increasing?

Wait — look again:

Sequence:
1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5

So:
- 1 → 1 time
- 2 → 2 times
- 3 → 2 times
- 4 → 3 times
- 5 → 4 times

Not matching.

Wait — maybe the repetition count follows: 1, 2, 2, 3, 4?

No.

Alternatively, think about how many times each number appears:

- 1: 1 time
- 2: 2 times
- 3: 2 times
- 4: 3 times
- 5: 4 times

Wait — maybe it's increasing by 1 every two numbers?

Or consider: the number of times n appears is n?

But:
- 1 appears 1 time → OK
- 2 appears 2 times → OK
- 3 appears 2 times → ✘ (should be 3)
- 4 appears 3 times → ✘ (should be 4)

No.

Wait — maybe it's the number of times a number appears equals its value minus something?

Another idea: look at when each number starts.

- 1: position 1
- 2: position 2
- 3: position 4
- 4: position 6
- 5: position 9

Wait — cumulative:

- 1: 1 item → total length: 1
- 2: 2 items → total: 1+2=3
- 3: 2 items → total: 3+2=5
- 4: 3 items → total: 5+3=8
- 5: 4 items → total: 8+4=12

So far, 12 terms.

Now, if the pattern is: number n appears n-1 times?

Check:
- 1: appears 1 time → n-1 = 0? No
- 2: appears 2 times → n-1 = 1? No

Wait — maybe the number of appearances is increasing: 1, 2, 2, 3, 4...

Wait — look at the frequency:

- 1 → 1 time
- 2 → 2 times
- 3 → 2 times
- 4 → 3 times
- 5 → 4 times

Wait — no clear rule.

Alternative idea: the number of times a number appears equals the previous number's frequency +1?

- 1: 1 time
- 2: 2 times → +1
- 3: 2 times → same
- 4: 3 times → +1
- 5: 4 times → +1

Still not clear.

Wait — maybe it's based on powers or digit counts?

Another idea: each number n appears n times, but shifted?

Try:
- 1 appears 1 time → [1]
- 2 appears 2 times → [2,2]
- 3 appears 3 times → [3,3,3] → but we have only two 3s → no

But we have:
- 1: 1
- 2: 2,2
- 3: 3,3
- 4: 4,4,4
- 5: 5,5,5,5

Ah! Now I see:

- 1: appears 1 time
- 2: appears 2 times
- 3: appears 2 times? Wait, no — we have only two 3s? But in the sequence: ...,3,3,4,...

Yes: positions 4 and 5: 3,3 → two 3s

But then 4: three 4s → positions 6,7,8: 4,4,4

5: four 5s → positions 9,10,11,12: 5,5,5,5

So:
- 1: 1 time
- 2: 2 times
- 3: 2 times
- 4: 3 times
- 5: 4 times

Now, the number of times n appears seems to be n-1 for n ≥ 2?

- 2: 2 times → n-1 = 1 → no
- 3: 2 times → n-1 = 2 → yes
- 4: 3 times → n-1 = 3 → yes
- 5: 4 times → n-1 = 4 → yes

But 2: should be 1 time? But it's 2 times.

Wait — maybe it's: for n ≥ 3, n appears (n-1) times, but 2 appears 2 times?

That doesn’t fit.

Wait — maybe it's based on how many times the number appears being equal to the number of digits in the sequence?

No.

Wait — let's count how many times each number appears:

- 1: 1 time
- 2: 2 times
- 3: 2 times
- 4: 3 times
- 5: 4 times

Wait — the number of times n appears is equal to the number of times (n-1) appeared?

- 1: 1
- 2: 2 → double?
- 3: 2 → same as 2
- 4: 3 → increase by 1
- 5: 4 → increase by 1

No.

Wait — maybe it's related to triangular numbers or something else.

Alternatively, look at run lengths:

- 1 → run of 1
- 2 → run of 2
- 3 → run of 2
- 4 → run of 3
- 5 → run of 4

So runs: 1, 2, 2, 3, 4

Next could be 5? So 6 appears 5 times?

But we need just the next number.

After five 5s, the next number should be 6, and since the run length is increasing: 1,2,2,3,4 → maybe next run is 5?

So six 6s?

But we only need the next term, which would be 6

But let's confirm.

Sequence:
1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, ...

So after 5,5,5,5 (four 5s), the next number is 6

And likely, 6 will appear five times.

So the next number is 6

✔ Answer: 6

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7. A, B, C, C, D, D, D, E, E, E, E, ___

Look at the pattern:

- A → 1 time
- B → 1 time
- C → 2 times
- D → 3 times
- E → 4 times

So the number of times a letter appears increases by 1 each time, starting from A.

But A:1, B:1, C:2, D:3, E:4

Wait — A and B both appear once, then C:2, D:3, E:4

So the pattern is: each letter appears (its position in alphabet) times?

- A: 1st → 1 time → yes
- B: 2nd → 2 times? But only 1 → no

Wait — maybe not.

Wait — count the occurrences:

- A: 1
- B: 1
- C: 2
- D: 3
- E: 4

So the run length is: 1,1,2,3,4

Next should be 5 times F?

So after E,E,E,E (four E's), next is F, and it will appear five times.

So the next letter is F

✔ Answer: F

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8. 4, 8, 12, 16, 20, 24, 28, 32, 36, ___

- Arithmetic sequence: increases by 4
- 36 + 4 = 40

✔ Answer: 40

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## ✔ Final Answers:

| # | Pattern | Next Item |
|---|--------|-----------|
| 1 | 10, 20, 30, ... | 90 |
| 2 | A, B, A, A, B, B, ... | A |
| 3 | 1, 3, 5, 7, ... | 21 |
| 4 | A, B, A, A, B, A, A, A, B, ... | A |
| 5 | 2, 4, 6, 8, ... | 22 |
| 6 | 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, ... | 6 |
| 7 | A, B, C, C, D, D, D, E, E, E, E, ... | F |
| 8 | 4, 8, 12, 16, ... | 40 |

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✔ Summary of Answers:

1. 90
2. A
3. 21
4. A
5. 22
6. 6
7. F
8. 40

Let me know if you'd like these explained further!