Geometric Sequences And Series

1. Geometric Sequences

Example: Find the General Element

i)  Recursively          ii)  Explicitly

A)  3, 6, 12, 24, 48, ...          B)  5, 15, 45, 135, ...

C)  -3, 30, -300, 3000, ...     D)  2, 2/3, 2/9, 2/27, ...

Solution

1. We see that to get to the next term, we only need to multiply the previous term by 2, hence

an = 2an-1,   a1 = 3

To find the nth term, we see that 3 is multiplied by two n - 1 times.  This is the same as multiplying by 2n-1.  Hence,

an = 3(2n-1)

2. We see that to get to the next term, we only need to multiply the previous term by 3, hence

an = 3an-1,   a1 = 5

To find the nth term, we see that 5 is multiplied by three n - 1 times.  This is the same as multiplying by 3n-1.  Hence,

an = 5(3n-1)

3. This sequence follows a similar pattern with a1 = -3 and a common ratio of -10.  Recursively we write

an = -10an-1,   a1 = -3

and explicitly

an = -3[(-10)n-1]

4. This sequence follows a similar pattern with a1 = 2 and a common ratio of 1/3.  Recursively we write

an = 1/3 an-1,   a1 = 2

and explicitly

an = 2[(1/3)n-1]

 Definition:   A geometric sequence is a sequence with a common quotient, r, ie.,            an = ran-1

 Theorem:   The general element of a geometric sequence is           an = a1rn-1

Proof:

(by induction)
For n = 1 it is trivial
Assume that the theorem is true for n = k - 1 then

ak - 1 = a1rk-2

Our goal is to show that

ak = a1rk-1

but

ak = rak-1 = r(a1rk-2) = a1rk-1

Hence by induction the theorem is true.

Example:

Suppose that the fifth term of a geometric sequence is 80 and the eleventh term is 5120.  Find the third term

Solution:

We have that

80 = a1r4

and

5120 =  a1r10

Dividing the equation, we get:

5120/80 = r6

or

r6 = 64

taking sixth roots we get

r = 2
or     r = -2

Hence

an = a12n - 1      or    an = a1(-2)n - 1

From this

80 = a124

so that

a1 = 5

We see that

an = 5(2n - 1)     or      an = 5((-2)n - 1)

Finally

a3 = 5(22) = 20    (note that 5((-2)2) = 20 also)

2. The Geometric Series:

 Theorem      S i = 1n  a1ri-1  = a1(1 - rn)/(1 - r)

Proof:

By induction

For n = 1

a1 = a1(1 - r1)/(1 - r) = a1

Now assume the theorem is true for n = k - 1, then

S i = 1k-1  a1ri-1  = a1(1 - rk-1)/(1 - r)

Our goal is to show that

S i = 1k  a1ri-1  = a1(1 - rk)/(1 - r)

The left hand side is

S i = 1k  a1ri-1  = (S i = 1k-1  a1ri-1) + a1rk-1

= a1(1 - rk-1)/(1 - r) + a1rk-1

= [a1/(1 - r)](1 - rk-1 + (1 - r)rk-1)

= [a1/(1 - r)](1 - rk-1 + rk-1 - rk)

= [a1/(1 - r)](1 - rk) = a1(1 - rk)/(1 - r)

By induction the theorem is true.

Example:

Find

S i = 120 [3(1/2)i]

Solution:

We see that

a1 = 3     and     r = 1/2

Hence the sum is

3(1 - (1/2)20)/(1 - 1/2) @ 5.999994

3. Application:  Annuities

I pay \$300 per month into an annuity till I am 65 years old (a total of 34 years).  The annuity earns 6% interest. How much will I get at age 65?

Solution:

We consider the process backwards.  The last month's payment will accrue no interest, the second to last month's payment will accrue 2 months of interest, the third to last month's payment will accrue 3 month's of interest, and so on.  Putting these all into the compound interest formula and adding them we get

300 + 300(1 + .06/12) + 300(1 + .06/12)2
+ 300(1 + .06/12)3 + ... + 300(1 + .06/12)407
(notice that there are (34)(12) = 408 payments and 0, ..., 407 are 408 numbers)

= Si=1408[300(1.005)i-1] = 300[(1 - 1.005408)/(1 - 1.005)]

= \$399,096.99