How to use this applet:

Dr. Medieval has hijacked a satellite system, he wants to take out the worlds communications system using a polymorphic stealth virus. Your mission, should you choose not to decline it, is to break the virus deployment sequence and disable the deployment subroutine.

Luckily the security system has been programmed by a bunch of first year programmers
from Trinity College, so the code is easy enough to infiltrate, however the sequence retrieval system is written by the same shower, so we can only get a certain number of sequences!

The deal is, we can get four terms of a geometric sequence, starting with the second term. You must find the next four terms before the clock hits zero.

Choose the level of difficulty, click start and when the sequence has been entered click diffuse!

Hint: To try again hit reload.

This applet will self-construct.

Notes on the maths used in the applet:

If you let (as most maths textbooks do) the first term be equal to a and the common ratio be equal to r the sequence formed is:

a, ar, ar², ar³, ar⁴,....ar^n-1 

The last term of the above sequence is known as the nth term and mathematically it is defined as

U_n = ar^n-1

In order to ascertain if a given geometric sequence is, in fact, geometric we must prove that U_n+1/U_n = C . Where C is a constant.

Example:

Say we were given x number of terms in a geometric sequence and were asked to find
more terms of the sequence. In order to accomplish this we must find the constant a and 
the common ratio, r. That is, in the sequence Un = 1,2, 4, x, y,...., to find x and y it is necessary to find a, (easy because it is always the first term) and your common ratio r. From this we can calculate x and y:

Using: U_n = ar^n-1

        U₁ = 1*k^(1-1)  (a = 1)    (k is irrelevant since x⁰ = 1)

        U₁ = 1*1

        U₂ = 1*2^(2-1) =  2         (therefore r = 2)

        U₃ = 1*2^(3-1) =  4 

        U₄ = 1*2^(4-1) =  8

Then x = 4, y = 8.

So, in order to calculate any terms in a geometric series we have to have the common ratio, r, or other terms of the sequence (i.e. a means of which to find r). Why not try
your hand at the little Java game below, it helps demonstrate the application of the above and is very good practice for quick sequence formation.

Another common example is when you are given the nth term of a sequence and asked to calculate a certain number of terms. 

Take the general term Un = 6(^-1/₂)n-1 and calculate the first four terms. Confirm this sequence is geometric.

         U₁ = 6(^-1/₂)^1-1 = 6

         U₂ = 6(^-1/₂)2-1 = -3

         U₃ = 6(^-1/₂)3-1 = ³/₂  

To confirm this sequence is a geometric one it is necessary to confirm Un+1/Un is a constant so:

6(^-1/₂)ⁿ / 6(^-1/₂)^n-1 = ^-1/₂

Geometric Series are formed when a geometric sequence is summed. 

ar + ar2 + ar3 + ........ + arn-1

Sn can easily be found by subtraction:

Sn = a +  ar + ar² + ar³ + ........ + ar^n-1

rSn =       ar + ar² + ar³ + ........ + ar^n-1 + arⁿ 

S_n(1-r) = a(1-rⁿ)

S_n = a(1-rⁿ) / (r - 1)   [when | r | < 1] and

S_n = a(rⁿ - 1) / (r -1)  [when | r | > 1]