Sequence
See how to find a sequence and a series
(arithmetic/geometric/other).
23 examples and their solutions.
Arithmetic Sequence
Definition
whose difference of the adjacent terms
are the same.
So, if you add +d,
you get the next term.
Formula
an = a + (n - 1)d
an: nth term
a: first term, a1
d: common difference
an: nth term
a: first term, a1
d: common difference
Example
1, 4, 7, 10, 13, ...
an = ?
Solution an = ?
an = 1 + (n - 1)⋅3
= 1 + 3n - 3
= 3n - 2
Close
Example
-2, 5, 12, 19, ...
ak = 551, k = ?
Solution ak = 551, k = ?
ak = -2 + (k - 1)⋅7
= -2 + 7k - 7
= 7k - 9 = 551 - [1]
7k = 560
k = 80
[1]
ak = 7k - 9
ak = 551
ak = 551
Close
Example
Arithmetic sequence: a8 = 5, a12 = 13
an = ?
Solution an = ?
a8 = a + 7d = 5
a12 = a + 11d = 13
a + 11d = 13
-a + 7d = 5
4d = 8 - [1]
d = 2
a + 7⋅2 = 5 - [2]
a + 14 = 5
a = -9
an = -9 + (n - 1)⋅2
= -9 + 2n - 2
= 2n - 11
a12 = a + 11d = 13
a + 11d = 13
-a + 7d = 5
4d = 8 - [1]
d = 2
a + 7⋅2 = 5 - [2]
a + 14 = 5
a = -9
an = -9 + (n - 1)⋅2
= -9 + 2n - 2
= 2n - 11
[2]
d = 2 → a + 7d = 5
Close
Arithmetic Sequence: Mean
Definition
Arithmetic Means:
the middle terms that form an arithmetic sequence
with the first and the last term.
the middle terms that form an arithmetic sequence
with the first and the last term.
Example
Find the three arithmetic means between 7 and 23.
Solution =
7 + 4d
7 + 4d = 23
4d = 16
d = 4
7 + 4 = 11
11 + 4 = 15
15 + 4 = 19 - [2]
11, 15, 19
[1]
Write 7 □ □ □ 23.
(□ □ □: three arithmetic means)
This is an arithmetic sequence.
So write the common difference d
between the terms.
(□ □ □: three arithmetic means)
This is an arithmetic sequence.
So write the common difference d
between the terms.
[2]
To find the three □,
+4 from the first term 7.
+4 from the first term 7.
Close
Arithmetic Series
Formula
Sn = n2[2a + (n - 1)d]
= n2[a + an]
Sn: a1 + a2 + a3 + ... + an
a: first term, a1
d: common difference
Series: partial sum of a sequence, Sn = n2[a + an]
Sn: a1 + a2 + a3 + ... + an
a: first term, a1
d: common difference
Example
Arithmetic sequence: a = 3, d = 5
S20 = ?
Solution S20 = ?
a = 3, d = 5, n = 20
S20 = 202(2⋅3 + 19⋅5)
= 10⋅(6 + 95)
= 10⋅101
= 1010
S20 = 202(2⋅3 + 19⋅5)
= 10⋅(6 + 95)
= 10⋅101
= 1010
Close
Example
-1 + 3 + 7 + 11 + ... + 123
Solution an = -1 + (n - 1)⋅4
= -1 + 4n - 4
= 4n - 5
4n - 5 = 123 - [1]
4n = 128
n = 32
S32 = 322(2⋅(-1) + 31⋅4)
= 16⋅(-2 + 124)
= 16⋅122
= 1952
[1]
an = 4n - 5
an = 123
an = 123
Close
Example
Sn = n2 + 2n
an = ?
Solution an = ?
1) n = 2, 3, 4, ...
an = Sn - Sn - 1 - [1]
= [n2 + 2n] - [(n - 1)2 + 2(n - 1)]
= n2 + 2n - [n2 - 2n + 1 + 2n - 2]
= n2 + 2n - [n2 - 1]
= n2 + 2n - n2 + 1
= 2n + 1
an = 2n + 1 (n = 2, 3, 4, ...)
2) n = 1
S1 = 12 + 2⋅1
= 1 + 2
= 3
a1 = 2⋅1 + 1 - [2]
= 2 + 1
= 3
S1 = a1 = 3
∴ an = 2n + 1 (n = 1) - [3]
∴ an = 2n + 1 (n = 1, 2, 3, 4, ...)
an = Sn - Sn - 1 - [1]
= [n2 + 2n] - [(n - 1)2 + 2(n - 1)]
= n2 + 2n - [n2 - 2n + 1 + 2n - 2]
= n2 + 2n - [n2 - 1]
= n2 + 2n - n2 + 1
= 2n + 1
an = 2n + 1 (n = 2, 3, 4, ...)
2) n = 1
S1 = 12 + 2⋅1
= 1 + 2
= 3
a1 = 2⋅1 + 1 - [2]
= 2 + 1
= 3
S1 = a1 = 3
∴ an = 2n + 1 (n = 1) - [3]
∴ an = 2n + 1 (n = 1, 2, 3, 4, ...)
[1]
Sn = a1 + a2 + a3 + ... + an - 1 + an
Sn - 1 = a1 + a2 + a3 + ... + an - 1
Sn - Sn - 1 = an
Sn - 1 = a1 + a2 + a3 + ... + an - 1
Sn - Sn - 1 = an
[2]
a1 from an = 2n + 1 (found in case 1)
[3]
If S1 ≠ a1,
the real a1 is S1.
(not a1 from an found in case 1)
the real a1 is S1.
(not a1 from an found in case 1)
Close
Sigma (Math)
Definition
∑k = 1nak = a1 + a2 + a3 + ... + an
Sigma, ∑, is a wayto write the sum of a sequence (series).
∑k = 1n ak is read as
[the sum of ak as k goes from 1 to n].
Sigma (Math)
Geometric Sequence
Definition
whose ratio of the adjacent terms
are the same.
So, if you multiply ×r,
you get the next term.
Formula
an = arn - 1
an: nth term
a: first term, a1
r: common ratio
an: nth term
a: first term, a1
r: common ratio
Example
2, 6, 18, 54, 162, ...
an = ?
Solution an = ?
an = 2⋅3n - 1
Close
Example
320, 160, 80, 40, ...
ak = 58, k = ?
Solution ak = 58, k = ?
ak = 320⋅(12)k - 1
= 26⋅5⋅2-k + 1
= 2-k + 7⋅5 = 58- [1]
2-k + 7⋅5 = 5⋅2-3
2-k + 7 = 2-3
-k + 7 = -3
k - 7 = 3
k = 10
[1]
ak = 2-k + 7⋅5
ak = 5/8
ak = 5/8
Close
Example
Geometric sequence: a2 = -6, a5 = 48
an = ?
Solution an = ?
a2 = ar = -6
a5 = ar4 = 48
ar4ar = 48-6
r3 = -8
r = -2
a⋅(-2) = -6 - [1]
a = 3
an = 3⋅(-2)n - 1
a5 = ar4 = 48
ar4ar = 48-6
r3 = -8
r = -2
a⋅(-2) = -6 - [1]
a = 3
an = 3⋅(-2)n - 1
[1]
r = -2 → ar = -6
Close
Geometric Sequence: Mean
Definition
Geometric Means:
the middle terms that form a geometric sequence
with the first and the last term.
the middle terms that form a geometric sequence
with the first and the last term.
Example
Find the four geometric means between 6 and 192.
Solution =
6⋅r5
6r5 = 192
r5 = 32
r = 2 - [2]
6⋅2 = 12
12⋅2 = 24
24⋅2 = 48
48⋅2 = 96 - [3]
12, 24, 48, 96
[1]
Write 6 □ □ □ □ 192.
(□ □ □ □: four geometric means)
This is a geometric sequence.
So write the common ratio r
between the terms.
(□ □ □ □: four geometric means)
This is a geometric sequence.
So write the common ratio r
between the terms.
[2]
[3]
To find the four □,
×2 from the first term 6.
×2 from the first term 6.
Close
Example
Find the three geometric means between 5 and 405.
Solution =
5⋅r4
5r4 = 405
r4 = 81
r = ±3
1) r = 3
5⋅3 = 15
15⋅3 = 45
45⋅3 = 135
2) r = -3
5⋅(-3) = -15
-15⋅(-3) = 45
45⋅(-3) = -135
±15, 45, ±135
Close
Geometric Series
Formula
Sn = a(rn - 1)r - 1
= a(1 - rn)1 - r
Sn: a1 + a2 + a3 + ... + an
a: first term, a1
r: common ratio
= a(1 - rn)1 - r
Sn: a1 + a2 + a3 + ... + an
a: first term, a1
r: common ratio
Example
Geometric sequence: a = 3, r = 2
S5 = ?
Solution S5 = ?
a = 3, r = 2, n = 7
S7 = 3(27 - 1)2 - 1
= 3⋅(128 - 1)
= 3⋅127
= 381
S7 = 3(27 - 1)2 - 1
= 3⋅(128 - 1)
= 3⋅127
= 381
Close
Example
Geometric sequence: a1 = 4, a3 = 36
S5 = ?
Solution S5 = ?
a1 = a = 4
a3 = 4r2 = 36
r2 = 9
r = ±3
1) r = 3
S5 = 4(35 - 1)3 - 1
= 4(243 - 1)2
= 2⋅242
= 484
2) r = -3
S5 = 4((-3)5 - 1)-3 - 1
= 4(-243 - 1)-4
= -(-244)
= 244
S5 = 484, 244
a3 = 4r2 = 36
r2 = 9
r = ±3
1) r = 3
S5 = 4(35 - 1)3 - 1
= 4(243 - 1)2
= 2⋅242
= 484
2) r = -3
S5 = 4((-3)5 - 1)-3 - 1
= 4(-243 - 1)-4
= -(-244)
= 244
S5 = 484, 244
Close
Example
∑k = 14 5⋅(23)k
Solution ∑k = 14 5⋅(23)k - [1]
a1 = 5⋅(23)1 - [2]
= 103
r = 23 - [3]
S4 = 103(1 - (23)4)1 - 23
= 103(1 - 1681)13
= 10⋅(8181 - 1681)
= 10⋅6581
= 65081
a1 = 5⋅(23)1 - [2]
= 103
r = 23 - [3]
S4 = 103(1 - (23)4)1 - 23
= 103(1 - 1681)13
= 10⋅(8181 - 1681)
= 10⋅6581
= 65081
[1]
[2]
ak = 5⋅(2/3)k
→ a1 = 5⋅(2/3)1
→ a1 = 5⋅(2/3)1
[3]
ak = 5⋅(2/3)k
→ r = (2/3)
→ r = (2/3)
Close
Infinite Geometric Series
Infinite Geometric Sequence
Formula
S = a1 - r (-1 < r < 1)
S: a1 + a2 + a3 + ...
a: first term, a1
r: common ratio
S: a1 + a2 + a3 + ...
a: first term, a1
r: common ratio
Example
1 + 12 + 14 + 18 + ...
Solution Example
10 - 203 + 409 - 8027 + ...
Solution S = 101 - (-23)
= 101 + 23
= 1033 + 23
= 10153
= 10⋅31⋅5
= 2⋅3
= 6
Close
Example
∑n = 0∞ 8⋅(17)n
Solution ∑n = 0∞ 8⋅(17)n - [1]
a1 = 8⋅(17)0 - [2]
= 8⋅1
= 8
r = 17 - [3]
S = 81 - 17
= 877 - 17
= 8167
= 4137
= 4⋅71⋅3
= 283
a1 = 8⋅(17)0 - [2]
= 8⋅1
= 8
r = 17 - [3]
S = 81 - 17
= 877 - 17
= 8167
= 4137
= 4⋅71⋅3
= 283
[1]
[2]
n: 0, 1, 2, ...
→ an = 8⋅(1/7)n - 1
→ a1 = 8⋅(1/7)0
→ an = 8⋅(1/7)n - 1
→ a1 = 8⋅(1/7)0
[3]
an = 8⋅(1/7)n - 1
→ r = (1/7)
→ r = (1/7)
Close
Repeating Decimal
Definition
0.123 = 0.1232323...
A repeating decimal is a decimalthat has a repeating part.
The numbers under the bar is the repeating part.
A repeating decimal is a rational number.
So you can change a repeating decimal
to a fraction.
Example
0.123 → fraction?
Solution 0.123 = 0.1232323...
100⋅0.123 = 12.3232323... - [1]
-0.123 = 0.1232323...
99⋅0.123 = 12.2
0.123 = 12.299
= 122990
= 61495
100⋅0.123 = 12.3232323... - [1]
-0.123 = 0.1232323...
99⋅0.123 = 12.2
0.123 = 12.299
= 122990
= 61495
[1]
0.123 = 0.1232323...
2 digits are under the bar.
So ×102 = ×100 both sides.
2 digits are under the bar.
So ×102 = ×100 both sides.
Close
Recursive Formula
Example
a1 = 4, an + 1 = an + 6
a1 ~ a4 = ?
Solution a1 ~ a4 = ?
a1 = 4
an + 1 = an + 6
a2 = a1 + 6
= 4 + 6
= 10
a3 = a2 + 6
= 10 + 6
= 16
a4 = a3 + 6
= 16 + 6
= 22
4, 10, 16, 22
an + 1 = an + 6
a2 = a1 + 6
= 4 + 6
= 10
a3 = a2 + 6
= 10 + 6
= 16
a4 = a3 + 6
= 16 + 6
= 22
4, 10, 16, 22
Close
Example
a1 = -2, an + 1 = an + 3n
a1 ~ a4 = ?
Solution a1 ~ a4 = ?
a1 = -2
an + 1 = an + 3n
a2 = a1 + 3⋅1
= -2 + 3
= 1
a3 = a2 + 3⋅2
= 1 + 6
= 7
a4 = a3 + 3⋅3
= 7 + 9
= 16
-2, 1, 7, 16
an + 1 = an + 3n
a2 = a1 + 3⋅1
= -2 + 3
= 1
a3 = a2 + 3⋅2
= 1 + 6
= 7
a4 = a3 + 3⋅3
= 7 + 9
= 16
-2, 1, 7, 16
Close
Example
a1 = 1, a2 = 1, an + 2 = an + an + 1
a1 ~ a7 = ?
This sequence is the Fibonacci Sequence. a1 ~ a7 = ?
Solution
a1 = 1
a2 = 1
an + 2 = an + an + 1
a3 = a1 + a2
= 1 + 1
= 2
a4 = a2 + a3
= 1 + 2
= 3
a5 = a3 + a4
= 2 + 3
= 5
a6 = a4 + a5
= 3 + 5
= 8
a7 = a5 + a6
= 5 + 8
= 13
1, 1, 2, 3, 5, 8, 13
a2 = 1
an + 2 = an + an + 1
a3 = a1 + a2
= 1 + 1
= 2
a4 = a2 + a3
= 1 + 2
= 3
a5 = a3 + a4
= 2 + 3
= 5
a6 = a4 + a5
= 3 + 5
= 8
a7 = a5 + a6
= 5 + 8
= 13
1, 1, 2, 3, 5, 8, 13
Close
Mathematical Induction
Definition
1. Show that n = 1 is true.
2. Assume n = k is true.
3. Show that n = k + 1 is true.
A mathematical induction is a way2. Assume n = k is true.
3. Show that n = k + 1 is true.
to prove the given statement (given).
1. Show that (given) is true
when n = 1.
2. Assume that (given) is true
when n = k.
3. Use the (given) when n = k
to show that (given) is true
when n = k + 1.
Then, just like a recursive formula,
(given) is true.
(given) is true when n = 1.
→ (given) is true when n = 1 + 1 = 2.
→ (given) is true when n = 2 + 1 = 3.
...
Example
Prove the given statement.
(n is a natural number.)
1 + 2 + 3 + ... + n = n(n + 1)2
Solution (n is a natural number.)
1 + 2 + 3 + ... + n = n(n + 1)2
1) n = 1
1 = 1⋅22
= 1 ( o )
2) n = k
Assume that below is true.
1 + 2 + 3 + ... + k = k(k + 1)2
3) n = k + 1
1 + 2 + 3 + ... + k + k + 1 = k(k + 1)2 + k + 1 - [1]
= k(k + 1)2 + 2(k + 1)2
= (k + 1)(k + 2)2
= (k + 1)((k + 1) + 1)2 ( o )
∴ The given statement is true.
1 = 1⋅22
= 1 ( o )
2) n = k
Assume that below is true.
1 + 2 + 3 + ... + k = k(k + 1)2
3) n = k + 1
1 + 2 + 3 + ... + k + k + 1 = k(k + 1)2 + k + 1 - [1]
= k(k + 1)2 + 2(k + 1)2
= (k + 1)(k + 2)2
= (k + 1)((k + 1) + 1)2 ( o )
∴ The given statement is true.
[1]
1 + 2 + 3 + ... + k = k(k + 1)/2
+(k + 1) both sides.
Then make the right side
(k + 1)((k + 1) + 1)/2.
+(k + 1) both sides.
Then make the right side
(k + 1)((k + 1) + 1)/2.
Close