Continuing Sequences from Patterns Worksheet | PDF printable ... - Free Printable

Let's solve each problem step by step, analyzing the patterns and answering the questions.

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1) Squares in a Row

Pattern:
Each term adds one more square to the row. Each square has 4 sticks, but shared sides reduce total count.

- Term 1: 1 square → 4 sticks
- Term 2: 2 squares → 4 + 3 = 7 sticks (shared side)
- Term 3: 3 squares → 7 + 3 = 10 sticks
- So pattern: Each new square adds 3 sticks.

General formula:
Sticks = 3n + 1
(For n=1: 3(1)+1 = 4 ✔️)

6th term:
3(6) + 1 = 18 + 1 = 19 sticks

✔ Answer: 19 sticks

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2) Houses (Triangle on Top of Square)

Each "house" consists of:
- A square (4 sticks)
- A triangle on top (2 sticks), sharing the top side with square

So each house uses: 4 (square) + 2 (triangle) = 6 sticks

But wait — look at the figures:

- Term 1: One house → 5 sticks? Let’s count carefully.

Actually, observe:
- First house: square (4) + triangle (2) = 6 sticks? But they’re connected.

Wait — better to count actual lines.

Looking closely:
- House 1: 5 sticks (square base: 4, roof: 1 shared side, so triangle needs only 2 sticks? Wait no.)

Actually, let's draw it:

- Square: 4 sticks
- Triangle on top: 2 additional sticks (since base is shared)

So each house = 4 + 2 = 6 sticks

Now check if they share anything between houses?

No — each house is separate. So:
- Term 1: 6 sticks
- Term 2: 6 × 2 = 12
- Term 3: 18
- So linear: 6n

5th term: 6 × 5 = 30 sticks

✔ Answer: 30 sticks

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3) L-shaped Pattern

Each term adds a new square to the right end.

Let’s count sticks:

- Term 1: 3 squares in L shape → 3×4 = 12, but shared edges reduce.
- Actually: 3 squares arranged as 2 vertical, 1 horizontal at bottom right.
- Count edges:
- Vertical column: 3 squares → 3 rows of 2 sticks each (top and bottom), plus 3 verticals → but shared.

Better way: Count all outer edges.

Alternatively, count number of sticks used per figure.

Term 1:
- 3 squares: each has 4 sides, but shared sides are counted once.
- Shared: 2 internal edges (between squares)
- Total sticks = 3×4 – 2×2 = 12 – 4 = 8? No — not correct.

Wait, actually, each stick is a line segment.

Let’s count directly:

Term 1:
- Left column: 2 squares stacked → 3 vertical lines (left, middle, right) and 3 horizontal lines (top, middle, bottom). But wait — no.

Standard method: each square uses 4 sticks, but shared edges are subtracted.

But easier: just count visible sticks.

Term 1:
- Bottom-left square: 4 sticks
- Top-left square: shares bottom with first → adds 3
- Right square: shares left with bottom-left → adds 3
Total: 4 + 3 + 3 = 10 sticks

Term 2:
- Adds another square to the right of the last one
- New square: shares left edge → adds 3 sticks
- So from 10 → 13

Term 3: 13 + 3 = 16

Term 4: 16 + 3 = 19

So pattern: increases by 3 each time.

Start: 10, 13, 16, 19,...

So general formula:
Sticks = 3n + 7
Check:
n=1: 3+7=10 ✔️
n=2: 6+7=13 ✔️
Yes.

We want: 3n + 7 = 35
→ 3n = 28 → n = 28/3 ≈ 9.33 → Not integer.

Wait — that can’t be.

Wait — maybe I miscounted.

Let me re-count Term 1:

L-shape: 2 squares vertically, 1 square to the right of bottom one.

Squares:
- A (bottom-left): 4 sticks
- B (top-left): shares bottom edge with A → adds 3 sticks (top, left, right)
- C (bottom-right): shares left edge with A → adds 3 sticks (top, right, bottom)

Total: 4 + 3 + 3 = 10 ✔️

Term 2: Add D to the right of C → shares left edge → adds 3 → total 13

Term 3: add E → 16

Term 4: 19

So sequence: 10, 13, 16, 19,...

So: Sticks = 3n + 7

Set 3n + 7 = 35 → 3n = 28 → n = 9.33... → Not possible.

But 35 is not in the sequence?

Wait — maybe my formula is off.

Wait: n=1: 10 → 3(1)+7 = 10 ✔️
n=2: 13 → 3(2)+7 = 13 ✔️
n=3: 16 → 3(3)+7 = 16 ✔️
n=4: 19 → 3(4)+7 = 19 ✔️

So yes, 3n + 7

Now, 3n + 7 = 35 → 3n = 28 → n = 9.33 → not integer

So no term uses exactly 35 sticks?

But question says “Which term is made with 35 sticks?”

Hmm. Maybe I made an error.

Wait — perhaps the pattern is different.

Let’s count sticks again, but think differently.

Maybe the figures are built with unit squares, and each square uses 4 sticks, but shared edges are shared.

So for Term 1: 3 squares

- Number of sticks = total edges minus shared ones

Each square has 4 edges → 3×4 = 12

Shared edges:
- Between A and B: 1 shared
- Between A and C: 1 shared
- Total shared: 2

Each shared edge saves 1 stick (since both squares would have counted it)

So total sticks = 12 – 2 = 10 ✔️

Term 2: 4 squares → 4×4 = 16 edges
Shared edges:
- A-B, A-C, C-D → 3 shared
→ 16 – 3 = 13 ✔️

Term 3: 5 squares → 20 edges
Shared: A-B, A-C, C-D, D-E → 4 shared → 20–4=16

So yes: sticks = 4n + 2? Wait:

n=1: 3 squares → 10 sticks
n=2: 4 squares → 13
n=3: 5 squares → 16
n=4: 6 squares → 19

Number of squares = n + 2

Because:
- Term 1: 3 squares
- Term 2: 4 squares
- So squares = n + 2

Each square contributes 4 sticks, but shared edges reduce.

Number of shared edges: (n + 1) ? Because each new square after the first adds one shared edge.

From above:
- n=1: 2 shared edges
- n=2: 3 shared edges
- n=3: 4 shared edges
- So shared = n + 1

Total sticks = 4 × (n + 2) – 2 × (n + 1)
= 4n + 8 – 2n – 2 = 2n + 6

Wait! Try this:

n=1: 2(1)+6 = 8 ✘ but we have 10

No.

Wait: total edges = 4 × (number of squares) = 4(n+2)
Shared edges = number of internal connections = (n+1) ? For n=1: 2 squares connected? No.

Wait — in Term 1: 3 squares, 2 shared edges
Term 2: 4 squares, 3 shared edges
So shared edges = n + 1

So total sticks = 4 × (n+2) – 2 × (n+1) = 4n + 8 – 2n – 2 = 2n + 6

But n=1: 2+6=8 ≠ 10 → wrong

Ah! The formula is: total sticks = total edges – shared edges

But each shared edge is counted twice, so we subtract once per shared edge.

So total sticks = (4 × #squares) – (#shared edges)

#squares = n + 2
#shared edges = n + 1

So: sticks = 4(n+2) – (n+1) = 4n + 8 – n – 1 = 3n + 7

Yes! That matches earlier.

So sticks = 3n + 7

Set 3n + 7 = 35 → 3n = 28 → n = 9.33 → not integer

So no term uses exactly 35 sticks.

But the question asks “Which term is made with 35 sticks?” implying there is one.

Wait — maybe I miscounted the number of squares.

Look at the figures:

- Term 1: 3 squares
- Term 2: 4 squares
- Term 3: 5 squares
- Term 4: 6 squares

Yes, so squares = n + 2

And sticks = 3n + 7

So:

n=1: 10
n=2: 13
n=3: 16
n=4: 19
n=5: 22
n=6: 25
n=7: 28
n=8: 31
n=9: 34
n=10: 37

So 35 is not in the sequence.

Closest: 34 (n=9), 37 (n=10)

So no term uses 35 sticks.

But maybe the pattern is different.

Wait — perhaps the fourth term is already drawn, and we need to see how many sticks.

But the question is: Which term is made with 35 sticks?

Since 35 is not in the sequence, answer is: None

But let's double-check.

Alternative idea: Maybe the pattern grows differently.

Wait — perhaps the number of sticks increases by 3 each time: 10, 13, 16, 19, 22, 25, 28, 31, 34, 37...

So 35 is not in the sequence.

So answer: There is no term made with exactly 35 sticks.

But let's see if the pattern is different.

Wait — maybe the fourth term is not 6 squares?

Looking at the image:

- Term 1: 3 squares
- Term 2: 4 squares
- Term 3: 5 squares
- Term 4: 6 squares

Yes.

So sticks: 10, 13, 16, 19, ...

So arithmetic sequence: d=3, a=10

So term n: a_n = 10 + (n-1)*3 = 3n + 7

Same as before.

So 3n + 7 = 35 → n = 28/3 → not integer.

✔ Answer: No term uses exactly 35 sticks. The closest are 34 (term 9) and 37 (term 10).

But since it asks "which term", maybe it's expecting "none".

Or perhaps I misread the pattern.

Wait — maybe the first term has 1 square? But no — it clearly shows 3 squares.

Wait — look again.

Term 1: two squares stacked, one attached to bottom-right — yes, 3 squares.

So I think my analysis is correct.

✔ Answer: No term uses exactly 35 sticks. It is not possible.

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4) Hexagons in a Row

Each hexagon has 6 sides.

When placed next to each other, they share one side.

So:
- Term 1: 6 sticks
- Term 2: 6 + 5 = 11 (second hexagon shares one side)
- Term 3: 11 + 5 = 16
- So: sticks = 5n + 1

Check:
n=1: 5+1=6 ✔️
n=2: 10+1=11 ✔️
n=3: 15+1=16 ✔️

Can a term use 51 sticks?

Set 5n + 1 = 51 → 5n = 50 → n = 10

Yes, the 10th term uses 51 sticks.

✔ Answer: Yes, the 10th term uses 51 sticks.

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5) Zigzag Pattern (like a fence)

Each unit is like a "V" or "M" shape.

Count sticks:

Term 1: 2 triangles? Or 2 sides?

It looks like two adjacent squares forming a zigzag.

Actually: each "unit" is a parallelogram-like shape made of 2 sticks?

Wait — better to count.

Term 1: 4 sticks? Looks like a "W" but smaller.

Wait — each shape is made of two connected squares forming a zigzag.

But let’s count:

Term 1: 4 sticks? No — it’s like two sides.

Wait — each "unit" is a rhombus made of 2 sticks?

No — looking closely:

Each "unit" appears to be a pair of sticks forming a V.

But in the pattern:

- Term 1: 2 V shapes? No — one "zig" shape.

Actually, each "segment" is a diamond made of 2 sticks? No.

Wait — it's like a series of connected chevrons.

Each chevron (like < >) is made of 2 sticks.

But they share vertices.

Wait — look:

Term 1: one chevron → 2 sticks? But it's drawn as two lines.

Actually, each chevron has 2 sticks, but when connected, they share a vertex.

But the pattern is:

- Term 1: 2 sticks
- Term 2: 4 sticks?
- Term 3: 6 sticks?

Wait — no — look at the drawing:

Term 1: one "M" shape? No — it's like a single zigzag: two lines forming a "V"

But it’s drawn with two sticks.

Term 2: two such V’s connected — but shared vertex.

So total sticks: 3? No — each V has 2 sticks, but when joined, they share a stick?

No — in the diagram, it looks like:

Term 1: 2 sticks (a V)
Term 2: 4 sticks (two Vs connected)
Term 3: 6 sticks
Term 4: 8 sticks

So pattern: 2, 4, 6, 8 → even numbers

So sticks = 2n

So 10th term: 2×10 = 20 sticks

✔ Answer: 20 sticks

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6) Cross Pattern with Gray Center

Each term has a gray rectangle and white tiles around.

Look:

- Term 1: gray 1×1, white: 4 tiles (one on each side)
- Term 2: gray 1×2, white: 6 tiles
- Term 3: gray 1×3, white: 8 tiles
- Term 4: gray 1×4, white: 10 tiles

So white tiles: 4, 6, 8, 10 → increasing by 2

So white tiles = 2n + 2

Check:
n=1: 2(1)+2=4 ✔️
n=2: 4+2=6 ✔️
n=3: 6+2=8 ✔️
n=4: 8+2=10 ✔️

So white tiles = 2n + 2

Set 2n + 2 = 19 → 2n = 17 → n = 8.5 → not integer

So no term has exactly 19 white tiles.

Also, white tiles are always even: 4,6,8,10,... → all even numbers

19 is odd → impossible

✔ Answer: No, because the number of white tiles is always even (2n+2), and 19 is odd.

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7) Dot Pattern

Each term has a vertical stack of dots and a horizontal row.

Term 1: 1 vertical, 2 horizontal → total 3 dots
Term 2: 2 vertical, 3 horizontal → 5 dots
Term 3: 3 vertical, 4 horizontal → 7 dots
Term 4: 4 vertical, 5 horizontal → 9 dots

So pattern: vertical = n, horizontal = n+1

Total dots = n + (n+1) = 2n + 1

So 12th term: 2(12) + 1 = 24 + 1 = 25 dots

✔ Answer: 25 dots

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8) Staircase Pattern

This pattern is different because:

- All previous patterns grow in a straight line (horizontal or vertical).
- This one grows diagonally, forming a staircase.
- Also, it uses rectangles (or rectangles), but each new term adds a rectangle at a diagonal position.
- Unlike others, it doesn't extend in one direction; it builds a diagonal line.
- Additionally, the number of units increases: 1, 2, 3, 4 — but arranged in a zigzag.

But key difference: It forms a diagonal structure, whereas others form linear or grid-based extensions.

✔ Answer: This pattern grows diagonally in a staircase fashion, unlike the others which grow horizontally or vertically in a straight line.

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✔ Final Answers:

1) 19 sticks
2) 30 sticks
3) No term uses exactly 35 sticks (closest: 34 or 37)
4) Yes, the 10th term uses 51 sticks
5) 20 sticks
6) No, because the number of white tiles is always even
7) 25 dots
8) It grows diagonally in a staircase pattern, unlike the linear growth of others.

Let me know if you'd like a visual explanation or help drawing the fourth terms!