A25b - Finding the nth term of a quadratic sequence - BossMaths.com - Free Printable

Let's solve each sequence step by step to find the 20th term.

We'll analyze the pattern in each sequence and derive a formula for the $ n $-th term, then compute $ a_{20} $.

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(a) 8, 13, 20, 29, 40, ...

Look at the differences between terms:

- $ 13 - 8 = 5 $
- $ 20 - 13 = 7 $
- $ 29 - 20 = 9 $
- $ 40 - 29 = 11 $

Differences: $ +5, +7, +9, +11 $ → increasing by 2 each time → quadratic sequence

So, this is a quadratic sequence of the form:
$$
a_n = an^2 + bn + c
$$

Let’s use the first three terms to solve for $ a, b, c $:

For $ n=1 $: $ a(1)^2 + b(1) + c = 8 $ → $ a + b + c = 8 $
For $ n=2 $: $ 4a + 2b + c = 13 $
For $ n=3 $: $ 9a + 3b + c = 20 $

Now solve:

1. $ a + b + c = 8 $
2. $ 4a + 2b + c = 13 $
3. $ 9a + 3b + c = 20 $

Subtract (1) from (2):
$ (4a + 2b + c) - (a + b + c) = 13 - 8 $ → $ 3a + b = 5 $ → (A)

Subtract (2) from (3):
$ (9a + 3b + c) - (4a + 2b + c) = 20 - 13 $ → $ 5a + b = 7 $ → (B)

Now subtract (A) from (B):
$ (5a + b) - (3a + b) = 7 - 5 $ → $ 2a = 2 $ → $ a = 1 $

Plug into (A): $ 3(1) + b = 5 $ → $ b = 2 $

From (1): $ 1 + 2 + c = 8 $ → $ c = 5 $

So,
$$
a_n = n^2 + 2n + 5
$$

Check:
- $ n=1 $: $ 1 + 2 + 5 = 8 $ ✔
- $ n=2 $: $ 4 + 4 + 5 = 13 $ ✔
- $ n=3 $: $ 9 + 6 + 5 = 20 $ ✔

Now $ a_{20} = 20^2 + 2(20) + 5 = 400 + 40 + 5 = \boxed{445} $

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(b) 0, 2, 6, 12, 20, ...

Differences:
- $ 2 - 0 = 2 $
- $ 6 - 2 = 4 $
- $ 12 - 6 = 6 $
- $ 20 - 12 = 8 $

Differences: $ +2, +4, +6, +8 $ → increasing by 2 → quadratic

Try $ a_n = an^2 + bn + c $

Use $ n=1 $: $ a + b + c = 0 $
$ n=2 $: $ 4a + 2b + c = 2 $
$ n=3 $: $ 9a + 3b + c = 6 $

Solve:

1. $ a + b + c = 0 $
2. $ 4a + 2b + c = 2 $
3. $ 9a + 3b + c = 6 $

(2)-(1): $ 3a + b = 2 $ → (A)
(3)-(2): $ 5a + b = 4 $ → (B)

(B)-(A): $ 2a = 2 $ → $ a = 1 $

Then (A): $ 3(1) + b = 2 $ → $ b = -1 $

From (1): $ 1 -1 + c = 0 $ → $ c = 0 $

So $ a_n = n^2 - n $

Check:
- $ n=1 $: $ 1 - 1 = 0 $ ✔
- $ n=2 $: $ 4 - 2 = 2 $ ✔
- $ n=3 $: $ 9 - 3 = 6 $ ✔

So $ a_{20} = 20^2 - 20 = 400 - 20 = \boxed{380} $

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(c) 2, 9, 18, 29, 42, ...

Differences:
- $ 9 - 2 = 7 $
- $ 18 - 9 = 9 $
- $ 29 - 18 = 11 $
- $ 42 - 29 = 13 $

Differences: $ +7, +9, +11, +13 $ → increase by 2 → quadratic

Try $ a_n = an^2 + bn + c $

Use:
- $ n=1 $: $ a + b + c = 2 $
- $ n=2 $: $ 4a + 2b + c = 9 $
- $ n=3 $: $ 9a + 3b + c = 18 $

(2)-(1): $ 3a + b = 7 $ → (A)
(3)-(2): $ 5a + b = 9 $ → (B)

(B)-(A): $ 2a = 2 $ → $ a = 1 $

Then (A): $ 3(1) + b = 7 $ → $ b = 4 $

From (1): $ 1 + 4 + c = 2 $ → $ c = -3 $

So $ a_n = n^2 + 4n - 3 $

Check:
- $ n=1 $: $ 1 + 4 - 3 = 2 $ ✔
- $ n=2 $: $ 4 + 8 - 3 = 9 $ ✔
- $ n=3 $: $ 9 + 12 - 3 = 18 $ ✔

So $ a_{20} = 20^2 + 4(20) - 3 = 400 + 80 - 3 = \boxed{477} $

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(d) 3, 8, 17, 30, 47, ...

Differences:
- $ 8 - 3 = 5 $
- $ 17 - 8 = 9 $
- $ 30 - 17 = 13 $
- $ 47 - 30 = 17 $

Differences: $ +5, +9, +13, +17 $ → increase by 4 → quadratic

Try $ a_n = an^2 + bn + c $

Use:
- $ n=1 $: $ a + b + c = 3 $
- $ n=2 $: $ 4a + 2b + c = 8 $
- $ n=3 $: $ 9a + 3b + c = 17 $

(2)-(1): $ 3a + b = 5 $ → (A)
(3)-(2): $ 5a + b = 9 $ → (B)

(B)-(A): $ 2a = 4 $ → $ a = 2 $

(A): $ 3(2) + b = 5 $ → $ b = -1 $

(1): $ 2 -1 + c = 3 $ → $ c = 2 $

So $ a_n = 2n^2 - n + 2 $

Check:
- $ n=1 $: $ 2 -1 + 2 = 3 $ ✔
- $ n=2 $: $ 8 - 2 + 2 = 8 $ ✔
- $ n=3 $: $ 18 - 3 + 2 = 17 $ ✔

So $ a_{20} = 2(400) - 20 + 2 = 800 - 20 + 2 = \boxed{782} $

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(e) 4, 10, 22, 40, 64, ...

Differences:
- $ 10 - 4 = 6 $
- $ 22 - 10 = 12 $
- $ 40 - 22 = 18 $
- $ 64 - 40 = 24 $

Differences: $ +6, +12, +18, +24 $ → increase by 6 → quadratic

Try $ a_n = an^2 + bn + c $

Use:
- $ n=1 $: $ a + b + c = 4 $
- $ n=2 $: $ 4a + 2b + c = 10 $
- $ n=3 $: $ 9a + 3b + c = 22 $

(2)-(1): $ 3a + b = 6 $ → (A)
(3)-(2): $ 5a + b = 12 $ → (B)

(B)-(A): $ 2a = 6 $ → $ a = 3 $

(A): $ 3(3) + b = 6 $ → $ b = -3 $

(1): $ 3 -3 + c = 4 $ → $ c = 4 $

So $ a_n = 3n^2 - 3n + 4 $

Check:
- $ n=1 $: $ 3 - 3 + 4 = 4 $ ✔
- $ n=2 $: $ 12 - 6 + 4 = 10 $ ✔
- $ n=3 $: $ 27 - 9 + 4 = 22 $ ✔

So $ a_{20} = 3(400) - 3(20) + 4 = 1200 - 60 + 4 = \boxed{1144} $

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(f) 3.5, 7, 11.5, 17, 23.5, ...

Differences:
- $ 7 - 3.5 = 3.5 $
- $ 11.5 - 7 = 4.5 $
- $ 17 - 11.5 = 5.5 $
- $ 23.5 - 17 = 6.5 $

Differences: $ +3.5, +4.5, +5.5, +6.5 $ → increase by 1 → quadratic

Try $ a_n = an^2 + bn + c $

Use:
- $ n=1 $: $ a + b + c = 3.5 $
- $ n=2 $: $ 4a + 2b + c = 7 $
- $ n=3 $: $ 9a + 3b + c = 11.5 $

(2)-(1): $ 3a + b = 3.5 $ → (A)
(3)-(2): $ 5a + b = 4.5 $ → (B)

(B)-(A): $ 2a = 1 $ → $ a = 0.5 $

(A): $ 3(0.5) + b = 3.5 $ → $ 1.5 + b = 3.5 $ → $ b = 2 $

(1): $ 0.5 + 2 + c = 3.5 $ → $ c = 1 $

So $ a_n = 0.5n^2 + 2n + 1 $

Check:
- $ n=1 $: $ 0.5 + 2 + 1 = 3.5 $ ✔
- $ n=2 $: $ 0.5(4) + 4 + 1 = 2 + 4 + 1 = 7 $ ✔
- $ n=3 $: $ 0.5(9) + 6 + 1 = 4.5 + 6 + 1 = 11.5 $ ✔

So $ a_{20} = 0.5(400) + 2(20) + 1 = 200 + 40 + 1 = \boxed{241} $

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✔ Final Answers:

(a) $ \boxed{445} $
(b) $ \boxed{380} $
(c) $ \boxed{477} $
(d) $ \boxed{782} $
(e) $ \boxed{1144} $
(f) $ \boxed{241} $

Let me know if you'd like visual explanations or graphs!