Geometric sequences and series

What are geometric sequences

Sometimes called geometric progressios (GPs), go up or down by the same multiple

'r'

common ratio.

Common ratio

r = tn / tn-1

or

r = tn / tn-1

Recursive definition

tn = r x tn-1

n>2

general nth term

tn = ar^n-1

Find 10th term of 2,6,18

a = 2, r =3

tn = ar^n-1

t19 = 2 x 3^9

= 39366

Find the common ratio of a = 18, t4 = 144.

a = 18

ar^3 = 144

18 x r^3 144

r^3 = 8

r = 2

sequence 36,18,9.

tn= 9/16

a = 36, 2= 1/2

tn = 36 x (1/2)^n-1 = 9/16

(1/2)^n-1 = 1/64

(1/2)^n-1 = (1/2)^6

n-1 = 6

n = 7

6th term of a GP is 10, and 6th is 80. Find a and r.

10 = ar^2

80 = ar^5

(2) / (1): r^3 = 8

r = 3

therefore r = 3 gives a 5/2

Compound interest.

Original amount invested and on any interestes subsequently generated. (for 8% interest : r=108% = 1.08)

Martha invests 2500 at 7 percent compounded anually. a) find value of investment after 5 years

b) how long until the investment is worth 10 000

q = 2500 r= 1.07

t6 = 2500 x 1.07^5

B) 2500 = 1.07^n-1 = 10000

1.07^n-1 = 4

n-1 = log 4/ log 1.07

Therefore, by the end of the 21st year the investment will be worth over 10 000. [tn is before the nth year]

Sum of Geometric Series

Sn=a(1-r^n)/1-r for -1

Find sum of the first 9 terms of the gp 1/3 , 1/9 , 1/27, 1/81

a = 1/3 , 2 = 1/9 / 1/3 = 1/3

S9 = 1/3(1-1/3^9)

/ 1-1/3

~0.499975

GP sequence 1,3,9... find how many terms must be added together to obtain a sum of 1093

a = 1, r =3, Sn=1093

sn= a(r^n-1)/r-1

1093=1(3^n-1)/3-1

2x1093=3^n-1

2187=3^n

n= log2187/log 3 (or use calculator)

n=7

Infinite Sum of a geometric sequence with -1

S=a/1-r

Infinite sum geometric: evaluate 6+2+2/3+...

a=6, r= 1/3

S= 6 divided by 1-1/3

= 6 over 2/3

=9

Recurring decimals in infinite sum of GP:

. .

Write 0.54 as a fraction.

0.54 = 0.54 +0.00054 + ...

a= 0.54, r =1/1000= 0.01

s=0.54 over 99/100

=54/100 x 100/99

= 54/99 = 6/11

or

s = 0.54/0.99

s = 54/99

= 6/11

Sn = a(1-r^n)/1-r as n =>infinite

= 1/2(1-1/2^n) divided by 1-1/2 as n => infinite

= 1-(1/22)^n as n => infinite

As n => infinite, (1/2)^n => 0

(Numbers less than 1 when squared get smaller, higher powers make it even smaller.)

=1-0=1