Geometric Sequence

A geometric series, just like an arithmetic series, is n amount of values in a geometric sequence summed. For example, the geometric series of the geometric sequence 3,9,27,243,729 is:

3 + 9 + 27 + 243 + 729

rSnFor geometric series', even more than arithmetic series', it is really inefficient to keep adding the term together. Hence, we have a general formula. This general formula might seem a little bit random, so I will derive it to show why it is like it is.

 When summing a geometric sequence you can do:

Sn =  u1+u1 × r1 + u1 × r2 +u1 × r3 + ...  +u1 × rn-1

If  Sn is multiplied by the ratio, we get

rSn = u1 × r +u1 × r2 +u1 × r3 + u1 × r4 + ... + u1 × rn-1 +u1 × rn  

By  subtracting  these  two  equations, you get:

Sn - rSn= u1 +u1 × rn.

of which you end up with

Sn=u1(1-rn)/(1-r) 

(You can also write it like this: Sn=u1(rn-1)/(r-1))

That is the general formula. 

If you are wondering why we multiply the first equation with r, and subtract them from each other, then the reason is because that is a way of finding the formula; it is something you can do in mathematics, to sort of find what you want. That way, you can take longer equations, and turn them into simpler ones, in an enclosed system.

To give an example, we have a geometric sequence where the first term is 5 and the ratio is 3. Now, find the sum of the first 14 terms.

Since we have all the numbers here, all we have to do is plug them into the general formula. So we do:

Sn=u1(1-rn)/(1-r) 

S14=5(1-314)/(1-3)

Popping out your calculator and inserting the numbers, gives you

S14= 11957420

As per usual, you will most likely be given other information than what is required by the general formula. If you have read the other topics we offer in Sequences and Series, you will know that all you have to do to find the different variables is to rearrange the general formula. 

if given the sum of the first nth terms and the ratio, you can find the first term of the series. If we rearrange the general formula, we get:

Sn=u1(1-rn)/(1-r) 

Sn(1-r)=u1(1-rn)

u1=(Sn(1-r))/(1-rn)

If we continue with the previous example, but act like we only know the sum of first 14 term and that the ratio is 3, we get

11957420 = =u1(1-314)/(1-3)

11957420 = u1(-4782968)/(-2)

u1=-23914840/-472968

u1= 5

We might also be given that the sum of a series, the ratio and the first term, and it would be in our best interest to find the term number. For this case, when rearranging the formula, it would be better to use the other formula mentioned: Sn=u1(rn-1)/(r-1). Using this reduces the amount of steps you have to take, and we end up with:

Sn(r-1)/u1= rn-1

(Sn(r-1)/u1)+1 = rn

n = logr (Sn(r-1)/u1)+1)

Logarithms has not been covered yet, but as long as you manage to put log with base r (whatever your ratio is), it should be fine. We can use our previous example here as well:

11957420 = 5(3n-1)/(3-1)

3n-1 = (11957420 × 2)/5

3n = 4782968 + 1

n =log3(4782969)

n = 14

The last variable, r, will soon be covered as well.