Introduction to Sequences - Kuta Software - Free Printable

Let's solve each problem step by step. We'll analyze the patterns in the sequences and find the required terms.

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Part 1: Find the next three terms in each sequence

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1) 1, –3, 9, –27, 81, ...

- This is a geometric sequence.
- Ratio: $ \frac{-3}{1} = -3 $, $ \frac{9}{-3} = -3 $, etc.
- Common ratio $ r = -3 $
- Next terms:
- $ 81 \times (-3) = -243 $
- $ -243 \times (-3) = 729 $
- $ 729 \times (-3) = -2187 $

✔ Next three terms: $ -243, 729, -2187 $

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2) 9, 109, 209, 309, 409, ...

- Differences: $ 109 - 9 = 100 $, $ 209 - 109 = 100 $, etc.
- Arithmetic sequence with common difference $ d = 100 $
- Next terms:
- $ 409 + 100 = 509 $
- $ 509 + 100 = 609 $
- $ 609 + 100 = 709 $

✔ Next three terms: $ 509, 609, 709 $

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3) 0, 3, 8, 15, 24, ...

- Look at pattern:
- $ 0 = 1^2 - 1 $
- $ 3 = 2^2 - 1 $
- $ 8 = 3^2 - 1 $
- $ 15 = 4^2 - 1 $
- $ 24 = 5^2 - 1 $
- So $ a_n = n^2 - 1 $
- Next:
- $ 6^2 - 1 = 36 - 1 = 35 $
- $ 7^2 - 1 = 49 - 1 = 48 $
- $ 8^2 - 1 = 64 - 1 = 63 $

✔ Next three terms: $ 35, 48, 63 $

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4) $ \frac{1}{2}, \frac{1}{2}, \frac{3}{8}, \frac{1}{4}, \frac{5}{32}, ... $

Let’s write all terms with denominator as power of 2:

- $ \frac{1}{2} = \frac{1}{2^1} $
- $ \frac{1}{2} = \frac{1}{2^1} $
- $ \frac{3}{8} = \frac{3}{2^3} $
- $ \frac{1}{4} = \frac{1}{2^2} $
- $ \frac{5}{32} = \frac{5}{2^5} $

Wait — numerators: 1, 1, 3, 1, 5

Denominators: $ 2^1, 2^1, 2^3, 2^2, 2^5 $

This seems irregular. Let's list indices:

| n | term | numerator | denominator |
|---|-------------|-----------|-------------|
| 1 | $ \frac{1}{2} $ | 1 | 2 |
| 2 | $ \frac{1}{2} $ | 1 | 2 |
| 3 | $ \frac{3}{8} $ | 3 | 8 = 2³ |
| 4 | $ \frac{1}{4} $ | 1 | 4 = 2² |
| 5 | $ \frac{5}{32} $| 5 | 32 = 2⁵ |

Numerators: 1, 1, 3, 1, 5 → odd numbers? But alternating?

Try to see if it's two interleaved sequences.

Odd positions (n=1,3,5): $ \frac{1}{2}, \frac{3}{8}, \frac{5}{32} $

- Numerators: 1, 3, 5 → odd numbers
- Denominators: 2, 8, 32 → $ 2^1, 2^3, 2^5 $ → powers of 2 with exponents 1,3,5

So for odd $ n = 2k-1 $:
$ a_{2k-1} = \frac{2k-1}{2^{2k-1}} $

Even positions (n=2,4): $ \frac{1}{2}, \frac{1}{4} $

- $ \frac{1}{2^1}, \frac{1}{2^2} $ → so $ a_{2k} = \frac{1}{2^k} $

So:
- n=6 (even): $ a_6 = \frac{1}{2^3} = \frac{1}{8} $
- n=7 (odd): k=4 → $ \frac{7}{2^7} = \frac{7}{128} $
- n=8 (even): $ a_8 = \frac{1}{2^4} = \frac{1}{16} $

✔ Next three terms: $ \frac{1}{8}, \frac{7}{128}, \frac{1}{16} $

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5) 4, 16, 36, 64, 100, ...

- These are squares: $ 2^2, 4^2, 6^2, 8^2, 10^2 $
- So $ a_n = (2n)^2 = 4n^2 $
- Next:
- $ 12^2 = 144 $
- $ 14^2 = 196 $
- $ 16^2 = 256 $

✔ Next three terms: $ 144, 196, 256 $

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6) 14, 34, 54, 74, 94, ...

- Arithmetic sequence: $ d = 20 $
- Next:
- $ 94 + 20 = 114 $
- $ 114 + 20 = 134 $
- $ 134 + 20 = 154 $

✔ Next three terms: $ 114, 134, 154 $

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7) $ 5, \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \frac{5}{16}, ... $

- Geometric sequence: $ a_n = 5 \cdot \left(\frac{1}{2}\right)^{n-1} $
- Ratio $ r = \frac{1}{2} $
- Next:
- $ \frac{5}{16} \cdot \frac{1}{2} = \frac{5}{32} $
- $ \frac{5}{64} $
- $ \frac{5}{128} $

✔ Next three terms: $ \frac{5}{32}, \frac{5}{64}, \frac{5}{128} $

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8) –9, 101, –999, 10001, –99999, ...

Look at absolute values:
- 9, 101, 999, 10001, 99999

Alternating signs: negative, positive, negative, positive, negative...

Pattern:
- $ -9 = -(10^1 - 1) $
- $ 101 = 10^2 + 1 $
- $ -999 = -(10^3 - 1) $
- $ 10001 = 10^4 + 1 $
- $ -99999 = -(10^5 - 1) $

So pattern: For odd $ n $: $ -(10^n - 1) $, even $ n $: $ 10^n + 1 $

So:
- $ a_6 $: even → $ 10^6 + 1 = 1000001 $
- $ a_7 $: odd → $ -(10^7 - 1) = -9999999 $
- $ a_8 $: even → $ 10^8 + 1 = 100000001 $

✔ Next three terms: $ 1000001, -9999999, 100000001 $

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Part 2: Find the tenth term in each sequence

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9) $ -1, \frac{2}{3}, \frac{7}{3}, 4, \frac{17}{3}, ... $

Write all terms with denominator 3:
- $ -1 = -\frac{3}{3} $
- $ \frac{2}{3} $
- $ \frac{7}{3} $
- $ 4 = \frac{12}{3} $
- $ \frac{17}{3} $

Numerators: $ -3, 2, 7, 12, 17 $

Differences: $ 2 - (-3) = 5 $, $ 7 - 2 = 5 $, $ 12 - 7 = 5 $, $ 17 - 12 = 5 $

Arithmetic sequence: $ a_n = \frac{-3 + (n-1)\cdot5}{3} $

So:
- $ a_{10} = \frac{-3 + 9\cdot5}{3} = \frac{-3 + 45}{3} = \frac{42}{3} = 14 $

✔ Tenth term: $ 14 $

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10) 7, 9, 12, 16, 21, ...

Differences:
- 9 - 7 = 2
- 12 - 9 = 3
- 16 - 12 = 4
- 21 - 16 = 5

So differences increase by 1: 2,3,4,5,...

So next differences: 6,7,8,9

So:
- $ a_6 = 21 + 6 = 27 $
- $ a_7 = 27 + 7 = 34 $
- $ a_8 = 34 + 8 = 42 $
- $ a_9 = 42 + 9 = 51 $
- $ a_{10} = 51 + 10 = 61 $

Alternatively, $ a_n = a_1 + \sum_{k=2}^{n} (k+1) $? Wait.

Actually, starting from $ a_1 = 7 $, and increments start at 2.

The increment at step $ n $ is $ n+1 $? Let's see:

- From $ a_1 $ to $ a_2 $: +2 → $ 1+1 = 2 $
- $ a_2 $ to $ a_3 $: +3 → $ 2+1 = 3 $
- So $ a_n = a_{n-1} + (n+1) $?

Wait: $ a_2 = a_1 + 2 $, $ a_3 = a_2 + 3 $, $ a_4 = a_3 + 4 $, ...

So $ a_n = a_{n-1} + n $

But $ a_2 = a_1 + 2 $, so increment is $ n $, not $ n+1 $

So recurrence: $ a_n = a_{n-1} + n $, $ a_1 = 7 $

Then:
- $ a_2 = 7 + 2 = 9 $
- $ a_3 = 9 + 3 = 12 $
- $ a_4 = 12 + 4 = 16 $
- $ a_5 = 16 + 5 = 21 $
- $ a_6 = 21 + 6 = 27 $
- $ a_7 = 27 + 7 = 34 $
- $ a_8 = 34 + 8 = 42 $
- $ a_9 = 42 + 9 = 51 $
- $ a_{10} = 51 + 10 = 61 $

✔ Tenth term: $ 61 $

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11) –2, –6, –18, –54, –162, ...

Geometric: $ r = 3 $
- $ a_n = -2 \cdot 3^{n-1} $
- $ a_{10} = -2 \cdot 3^9 = -2 \cdot 19683 = -39366 $

✔ Tenth term: $ -39366 $

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12) –23, –18, –8, –3, ...

Differences:
- $ -18 - (-23) = 5 $
- $ -8 - (-18) = 10 $
- $ -3 - (-8) = 5 $

So pattern: +5, +10, +5 → maybe alternating?

But only 4 terms. Let's assume pattern continues: +5, +10, +5, +10,...

So:
- $ a_5 = -3 + 10 = 7 $
- $ a_6 = 7 + 5 = 12 $
- $ a_7 = 12 + 10 = 22 $
- $ a_8 = 22 + 5 = 27 $
- $ a_9 = 27 + 10 = 37 $
- $ a_{10} = 37 + 5 = 42 $

But wait — this might be flawed. Let's check if it's quadratic or something.

Alternative idea: Maybe the pattern is based on parity.

But let’s try another approach: look at second differences.

First differences: 5, 10, 5

Second differences: 5, -5 → not constant.

Maybe not polynomial.

Alternatively, could it be that the pattern is:
- Odd positions: –23, –8, ...
- Even: –18, –3, ...

But only two per group.

Another idea: maybe the sequence is defined by a recurrence.

But perhaps it's simpler: the pattern of changes is +5, +10, +5, so next is +10?

So:
- $ a_5 = -3 + 10 = 7 $
- $ a_6 = 7 + 5 = 12 $
- $ a_7 = 12 + 10 = 22 $
- $ a_8 = 22 + 5 = 27 $
- $ a_9 = 27 + 10 = 37 $
- $ a_{10} = 37 + 5 = 42 $

Yes, alternates between +5 and +10, starting with +5.

So:
- $ a_1 = -23 $
- $ a_2 = -23 + 5 = -18 $
- $ a_3 = -18 + 10 = -8 $
- $ a_4 = -8 + 5 = -3 $
- $ a_5 = -3 + 10 = 7 $
- $ a_6 = 7 + 5 = 12 $
- $ a_7 = 12 + 10 = 22 $
- $ a_8 = 22 + 5 = 27 $
- $ a_9 = 27 + 10 = 37 $
- $ a_{10} = 37 + 5 = 42 $

✔ Tenth term: $ 42 $

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13) –4, 12, –36, 108, –324, ...

Geometric: $ r = -3 $
- $ a_n = -4 \cdot (-3)^{n-1} $
- $ a_{10} = -4 \cdot (-3)^9 = -4 \cdot (-19683) = 78732 $

✔ Tenth term: $ 78732 $

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14) –6, –2, 0, 1, $ \frac{3}{2} $, ...

Look at differences:
- $ -2 - (-6) = 4 $
- $ 0 - (-2) = 2 $
- $ 1 - 0 = 1 $
- $ \frac{3}{2} - 1 = \frac{1}{2} $

So differences: 4, 2, 1, 0.5 → halving each time!

So next differences: $ \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32} $

So:
- $ a_6 = \frac{3}{2} + \frac{1}{4} = \frac{7}{4} $
- $ a_7 = \frac{7}{4} + \frac{1}{8} = \frac{15}{8} $
- $ a_8 = \frac{15}{8} + \frac{1}{16} = \frac{31}{16} $
- $ a_9 = \frac{31}{16} + \frac{1}{32} = \frac{63}{32} $
- $ a_{10} = \frac{63}{32} + \frac{1}{64} = \frac{127}{64} $

Alternatively, notice:
- $ a_1 = -6 $
- $ a_2 = -2 = -6 + 4 $
- $ a_3 = 0 = -2 + 2 $
- $ a_4 = 1 = 0 + 1 $
- $ a_5 = 1.5 = 1 + 0.5 $
- $ a_6 = 1.75 $
- etc.

So $ a_n = a_{n-1} + \frac{4}{2^{n-2}} $ for $ n \geq 2 $? Let's see:

From $ a_2 $: add $ 4 = \frac{4}{1} $
$ a_3 $: add $ 2 = \frac{4}{2} $
$ a_4 $: add $ 1 = \frac{4}{4} $
$ a_5 $: add $ 0.5 = \frac{4}{8} $
So $ \text{add } \frac{4}{2^{n-2}} $

So $ a_n = a_{n-1} + \frac{4}{2^{n-2}} = a_{n-1} + \frac{2^{2}}{2^{n-2}} = a_{n-1} + 2^{4 - (n-2)} = a_{n-1} + 2^{6-n} $

But easier to just compute:

We have:
- $ a_5 = 1.5 $
- $ a_6 = 1.5 + 0.25 = 1.75 $
- $ a_7 = 1.75 + 0.125 = 1.875 $
- $ a_8 = 1.875 + 0.0625 = 1.9375 $
- $ a_9 = 1.9375 + 0.03125 = 1.96875 $
- $ a_{10} = 1.96875 + 0.015625 = 1.984375 $

As fraction:
- $ a_6 = \frac{7}{4} $
- $ a_7 = \frac{15}{8} $
- $ a_8 = \frac{31}{16} $
- $ a_9 = \frac{63}{32} $
- $ a_{10} = \frac{127}{64} $

✔ Tenth term: $ \frac{127}{64} $

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15) –28, 172, 372, 572, 772, ...

Differences:
- $ 172 - (-28) = 200 $
- $ 372 - 172 = 200 $
- $ 572 - 372 = 200 $
- $ 772 - 572 = 200 $

Arithmetic: $ d = 200 $

So $ a_n = -28 + (n-1)\cdot200 $

$ a_{10} = -28 + 9\cdot200 = -28 + 1800 = 1772 $

✔ Tenth term: $ 1772 $

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16) 37, 46, 55, 64, 73, ...

$ d = 9 $

$ a_{10} = 37 + 9\cdot9 = 37 + 81 = 118 $

✔ Tenth term: $ 118 $

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Part 3: Find the first four terms in each sequence

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17) $ a_n = \frac{2n + 1}{n^2} $

- $ a_1 = \frac{2(1)+1}{1^2} = \frac{3}{1} = 3 $
- $ a_2 = \frac{4+1}{4} = \frac{5}{4} $
- $ a_3 = \frac{6+1}{9} = \frac{7}{9} $
- $ a_4 = \frac{8+1}{16} = \frac{9}{16} $

✔ First four terms: $ 3, \frac{5}{4}, \frac{7}{9}, \frac{9}{16} $

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18) $ a_n = 3^{n-1} $

- $ a_1 = 3^0 = 1 $
- $ a_2 = 3^1 = 3 $
- $ a_3 = 3^2 = 9 $
- $ a_4 = 3^3 = 27 $

✔ First four terms: $ 1, 3, 9, 27 $

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19) $ a_n = n^2 + 1 $

- $ a_1 = 1 + 1 = 2 $
- $ a_2 = 4 + 1 = 5 $
- $ a_3 = 9 + 1 = 10 $
- $ a_4 = 16 + 1 = 17 $

✔ First four terms: $ 2, 5, 10, 17 $

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20) $ a_n = \frac{n^3}{2n + 1} $

- $ a_1 = \frac{1}{2+1} = \frac{1}{3} $
- $ a_2 = \frac{8}{4+1} = \frac{8}{5} $
- $ a_3 = \frac{27}{6+1} = \frac{27}{7} $
- $ a_4 = \frac{64}{8+1} = \frac{64}{9} $

✔ First four terms: $ \frac{1}{3}, \frac{8}{5}, \frac{27}{7}, \frac{64}{9} $

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✔ Final Answers Summary

#### Part 1: Next Three Terms
1) $ -243, 729, -2187 $
2) $ 509, 609, 709 $
3) $ 35, 48, 63 $
4) $ \frac{1}{8}, \frac{7}{128}, \frac{1}{16} $
5) $ 144, 196, 256 $
6) $ 114, 134, 154 $
7) $ \frac{5}{32}, \frac{5}{64}, \frac{5}{128} $
8) $ 1000001, -9999999, 100000001 $

#### Part 2: Tenth Term
9) $ 14 $
10) $ 61 $
11) $ -39366 $
12) $ 42 $
13) $ 78732 $
14) $ \frac{127}{64} $
15) $ 1772 $
16) $ 118 $

#### Part 3: First Four Terms
17) $ 3, \frac{5}{4}, \frac{7}{9}, \frac{9}{16} $
18) $ 1, 3, 9, 27 $
19) $ 2, 5, 10, 17 $
20) $ \frac{1}{3}, \frac{8}{5}, \frac{27}{7}, \frac{64}{9} $

Let me know if you'd like these formatted in a table or printed!