Math, Sequences and Series

Geometric Sequence

a geometric sequence is a sequence whose each term after the first is obtained by multiplying the preceding term by a nonzero constant called the common ratio.

What is Sequence

a sequence is a function whose domain is the finite set (1,2,3,.....,n) or the infinite set (1,23,....)

First Term

a1

Sum of Infinite Geometric Series

common ratio is less than 1 or greater than -1 but not 0.

Common Ratio

common ratio, r, can be determined by dividing any term in the sequence by the term that precedes it. Thus, in the geometric sequence {32, 16, 8, 4, 2,....} the common ratio is 1/2 since 16/32 = 1/2'

Find the Sum: 5 + 8 + 11 + ...... + 38

Explicit Formula: an = a1 + (n-1)d d = a2 - a1 = 8 -5 = 3 an = 5 + (n-1)3 an = 5 + 3n-3 an = 3n + 2 an = 38 --> 38 = 3n + 2 (transpose 2) 38-2 + 3n 36 = 3n 3 3 12 = n Sn = n/2 (a1 + an) --> S12 = 12/2 (a1 + a12) S12 = 6 (5 + 38) --> S12 = 6 (43) S12 = 258

Consider the geometric sequence 3, 6, 12, 24, 48, 96, ... What is the sum of the first 5 terms?

Let S5 = 3 + 6 + 12 + 24 + 48. multiplying both sides by the common ratio 2, we get 2S5 = 6 + 12 + 24 + 48 + 96 subtracting 2S5 from S5, we have S5 = 3 + 6 + 12 + 24 + 48 - (2S5 = 6 + 12 + 24 + 48 + 96) -S5 = 3 -96 -S5 = -93 S5 = 93

Insert 3 geometric means between 5 and 3125.

Let a1 = 5 and a5 = 3125. we will insert a2, a3, and a4. since a5 = a1r^4, then 3125 = 5r^4. (dividing 5 to 3125 beomes 625) solving for the value of r, we get 625 = r^4 or r = ±5. we obtained two values of r, so we have two geometric sequences. if r = 5, the geometric means are a2 = 5(-5)^1 = -25 a3 = 5(-5)^2 = 125 a4 = 5(-5)^3 = -625 thus, the sequence is 5, -25, 125, -625, 3125.

Sum of Arithmetic Sequence

Sn = n/2 (a1+an) where a1 - first term & an - last term

Sum of Arithmetic Sequence

Sn = n/2 (a1=an) or Sn = n/2 {a1 + [a1 + (n-1)d]} or Sn = n/2 [2a1 + (n-1)d]. 1). Find the sum of the first 10 terms of the arithmetic sequence {5,9,13,17,...} S10 = 10/2 [2(5) + (10-1)4] --> 5 [10 + 9(4)] --> 5 [10 + 36] --> 5 [46] S10 = 230 2). Find the sum of the first 20 terms of the arithmetic sequence {-2, -5, -8, -11, ..} S20 = 20/2 [2(=2) + (20-1)-3] --> 10 [-4 + (19)-3] --> 10 [-4 - 57] --> 10 --> 10 [-61] S20 = -610

Solve: a8 = 25, a14 = 43

1). a8 = a1+ (n-1)d --> a8 = a1 + (8-1)d = 25 --> a8 = a1 + 7d = 25 2). a14 = a1 + (n-1)d --> a14 = a1 + (14-1)d = 43 --> a14 = a1 + 13d = 43 FIND D a8 = a1 + 7d = 25 --> -a1 - 7d = -25 (transform into negative) a14 = a1 + 13d = 43 --> + a1 + 13d = 43 = 6d = 18 6 6 --> d = 3 FIND A1 a8 = a1 + 7d = 25 --> a1 + 7(3) = 25 --> a1 + 21 = 25 a14 = a1 + 13d = 43 --> a1 + 13(3) = 43 --> a1 + 39 = 43 a1 = 25-21 = 4 a1 = 43-39 = 4 --> a1 = 4

Arithmetic Sequence Definition

a sequence in which each term is found by adding the same number to the previous term. is a sequence where every term after the first is obtained by adding a constant called the common difference. a1 = a a2 = a + d a3 = a + 2d a4 = a + 3d

Try the method for sequence 81, 27, 9, 3, 1, ... and find the sum of the first 4 terms.

from the activiry, we can derive a formula for the sum of the first n terms, Sn, of a geometric sequence consider the sum of the first n terms of a geometric sequence: Sn = a1 + a1r + a1r^2 + .... + a1r^n-1 (equation 1) multiplying both sides of equation by the common ratio r, we get rSn = a1r + a1r^2 + a1r^3 + .... + a1r^n-1 + a1r^n (equation 2) subtracting equation 2 from equation 1, we get Sn = a1 + a1r + a1r^2 + .... + a1r^n-1 [equation 1] -(rSn = a1r + a1r^2 + a1r^3 + .... + a1r^n-1 + a1r^n) [equation 2] Sn - rSn = a1 -a1r^n factoring both sides of the resulting equation, we get Sn(1-r) = a1 (1-r^n) dividing both sides by 1-r, where 1-r ≠ 0, we get Sn = a1 (1-r^n)/1-r, r ≠ 1. Note that since an = a1r^n-1, if we multiply both sides by r we get an(r) = a1r^n-1(r) or anr = a1r^n Since Sn = a1-anr/1-r = a1 -a1r^n/1-r. then replacing a1r^n by anr, we have Sn = a1-anr/1-r, r ≠ 1. What if r = 1? If r=1, then the formula above is not applicable. Instead Sn = a1 + a1(1) + a1(1)^2 + .... + a1(!)^n-1

Geometric Means

inserting a certain number of terms between two given terms of a geometric sequence is an interesting activity in studying geometric sequences. we call the terms between any two given terms of a geometric sequence the geometric means.

Arithmetic Mean Formula

AM = a+b/2 Example: 85 and 95 --> 85+95/2 = 180/2 = 90

Sum of Arithmetic Sequence

In particular, if r = -1 the sum Sn simplifies to Sn = {0 if n is even, a1 if n is odd} What is the sum of the first 10 terms of 2 - 2 + 2 - 2 + .... ? Since r = -1 and n is even, then the sum is 0. What is the sum of the first 11 terns of 2-2+2-2+.... ? Since r = -1 and n is odd, then the sum if 2. What is the sum of the first five terms of 3,6,12,24,48,96,...? Since a1=3, r=2, and n=5, then the sum is sol 1: Sn = 3 - (1-2^5)/1-2 = 3(-31)/-1 = 93. Since a1=3, a5=48, and r=2. alt sol: Sn = 3-(48)(2)/1-2 = 3-96/-1 = -93/-1 = 93.

Common Difference Formula

a2 - a1

Arithmetic Sequence Formula

an = a1 + (n-1)d

Solve: a4 = 39, a7 = 57

an = a1 + (n-1)d --> a4 = a1 + (4-1)d = 39 --> a4 = a1 + 3d = 39 a7 = a1 + (7-1)d = 57 --> a7 = a1 + 6d = 57 FIND D: -a1 - 3d = -39 (transform into negative) + a1 + 6d = 57 3d = 18 3 3 d = 6 FIND A1: a1 + 3(6) = 39 --> a1 + 18 = 39 --> a1 = 39 - 18 = 21 a1 + 6(6) = 57 --> a1 + 36 = 57 --> a1 = 57 - 36 = 21 a1 = 21

Common Difference

d

Consider: 3,8,13,18,23,28,...

d = 5

Solve: 3,7,11,15,19,23,27,31,35,39. Find a). a10 b). a40

d = a2-a1 = 7-3 = 4 --> d=4 FIND THE EQUATION an = a1 + (n-1)d --> an = 3 + (n-1)4 --> an = 3 + 4n - 4 --> an = 4n-1 a). Find a10: a10 = 4(10)-1 = 39 b). Find a40: a40 = 4(40)-1 = 159

Geometric Sequence Formula

given the first term a1 and the common ratio r of a geometric sequence, the nth term of a geometric sequence is an = a1r^n-1 What is the 10th term of the geometric sequence 8, 4, 2, 1, ... ? Since r = 1/2 then a10 = 8(1/2)^10-1 --> a10 = 8(1/2)^9 --> a10 = 8(1/512) a10 = 1/64.

Insert 4 arithmetic means between 5 and 25.

since we are required to insert 4 items, then there will be 6 terms in all. Let a1 = 5, and a6 = 25. 5, a2, a3, a4, a5, 25. we need to get the common difference. let us use a6 = a + 5d to solve for d. substituting the given values for a6 and a1, we obtain 25 = 5 + 5d. so, d =4 5, 5+4, 9+4, 13+4, 17+4, 21+4 5, 9, 13, 17, 21, 25. the 4 arithmetic means between 5 and 25 are 9, 13, 17, and 21.

Arithmetic Mean Definition

the terms between any two nonconsecutive terms of an arithmetic sequence are known as arithmetic means.