# Advice 1: How to find the denominator of a geometric progression

According to the definition a geometric progression is a sequence of non-zero numbers, each of which is equal to the previous multiplied by a constant number (the denominator of the progression). While geometrically there should be no one scratch, otherwise the whole sequence of "reset" that contradicts the definition. To find the denominator, it is sufficient to know the values of its two neighbouring members. However, it is not always the conditions of the problem are so simple.
You will need
• calculator
Instruction
1
Divide any member of a progression on the previous one. If the value of the previous value is unknown or undefined (for example, the first member of the progression), then divide by any member of the sequence the value of the next member of the progression.
As a member of a geometric progression is not equal to zero when performing this operation should not cause problems.
2
Example.
Suppose we have a sequence of numbers:

10, 30, 90, 270...

You want to find the denominator of a geometric progression.
Solution:

Option 1. We will take a random value (e.g., 90) and divide it by the previous (30): 90/30=3.

Option 2. Take any member of a geometrical progression (for example, 10) and divide by it later (30): 30/10=3.

Answer: the denominator of the geometric progression 10, 30, 90, 270... is set to 3.
3
If the values of members of a geometrical progression is not clearly specified, but in the form of ratios, write and solve a system of equations.
Example.
The sum of the first and the fourth member of a geometric progression is equal to 400 (b1+b4=400), and the sum of the second and the fifth member is equal to 100 (b2+b5=100).

It is required to find the denominator of the progression.
Solution:

Write down the problem statement in the form of a system of equations:
b1+b4=400

b2+b5=100
From the definition of geometric progression implies that:

b2=b1*q

b4=b1*q^3

b5=b1*q^4, where q is the common designation of the denominator of a geometric progression.
Substituting in the equations the values of the members of the progression, we get:
b1+ b1*q^3=400

b1*q+ b1*q^4=100
After factorization, it turns out:
b1*(1+q^3)=400

b1*q(1+q^3)=100
Now divide the corresponding part of the second equation into the first:
[b1*q(1+q^3)] / [b1*(1+q^3)] = 100/400, where q=1/4.
4
If you know the amount of several members of a geometrical progression or the sum of all of the members of the decreasing geometric progression, to find the denominator of the progression use the appropriate formulas:
Sn = b1*(1-q^n)/(1-q), where Sn – sum of n first members of the geometric progression and
S = b1/(1-q), where S is the sum of infinitely decreasing geometric progression (sum of all members of the progression with denominator smaller units).
Example.

The first member of the decreasing geometric progression is equal to one and the sum of all of its members equal to two.

You want to determine the denominator of this progression.
Solution:

Substitute the data from the task into the formula. Work:
2=1/(1-q), where q=1/2.

# Advice 2: How to find the denominator progression

The progression is a sequence of numbers. Exponentially each subsequent term is obtained by multiplying the previous one by some number q, called the denominator of the progression.
Instruction
1
If you know two adjacent member geometric progression b(n+1) and b(n) to obtain the denominator, we need the number with a large index divided by the previous: q=b(n+1)/b(n). It follows from the definition of progression and its denominator. An important condition is the inequality to zero of the first member and the denominator of progression, or the progression is undefined.
2
So, between members of the progression are set by the following relationship: b2=b1•q, b3=b2•q, ... , b(n)=b(n-1) •q. According to the formula b(n)=b1•q^(n-1) can be computed for any member of a geometrical progression, in which well-known denominator q and the first member b1. Also, each of the members of a geometric progression modulo equal to the median of their neighboring members: |b(n)|=√[b(n-1)•b(n+1)], hence the progression got its name.
3
The analogue of the geometric progression is the simplest exponential function y=a^x, where argument x is in the exponent, a is some number. In this case, the denominator of the progression coincides with the first member and is equal to the number a. Under the value of the function y can be understood n-th member of progression, if the argument x is taken for a natural number n (the counter).
4
There is a formula for the sum of the first n terms of a geometric progression: S(n)=b1•(1-q^n)/(1-q). This formula is valid when q≠1. If q=1, then the sum of the first n terms is computed by the formula S(n)=n•b1. By the way, the progression is called increasing if q is greater than one and positive b1. When the denominator of the rate, not exceeding units, the progression will be called descending.
5
A special case of the geometric progression is infinite decreasing geometric progression (b..g.p.). The fact that the members of the decreasing geometric progression again and again will decrease but never reach zero. Despite this, it is possible to find the sum of all the members of this progression. It is defined by the formula S=b1/(1-q). The total number of members n is infinite.
6
In order to visualize how it can be folded an infinite number of numbers and not get infinity, bake a cake. Cut half of the cake. Then cut 1/2 of the half, and so on. Pieces that you get, represent not that other, as members of an infinitely decreasing geometric progression with the denominator of 1/2. If you add up all these pieces, you will receive the original cake.
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