Consider some series.
7 28 112 448 1792 ...
It is perfectly clear that the value of any of its elements is four times greater than the previous one. Hence, this series is a progression.
Geometric progression called an infinite sequence of numbers, the main feature of which is that the next number is obtained from the previous one by multiplying by some specific number. This is expressed by the following formula.
a z +1 = a z · q, where z is the number of the selected element.
Accordingly, z ∈ N.
The period when the school is studying geometric progression - grade 9. Examples will help to understand the concept:
0.25 0.125 0.0625 ...
18 6 2 ...
Based on this formula, the denominator of progression can be found as follows:
Neither q nor b z can be zero. Also, each of the elements of the numerical series of the progression should not be zero.
Accordingly, in order to find out the next number in a series, you need to multiply the latter by q.
To set this progression, you must specify its first element and denominator. After that, it is possible to find any of the following members and their sums.
Varieties
Depending on q and a 1, this progression is divided into several types:
- If both a 1 and q are greater than unity, then such a sequence is a geometric progression increasing with each next element. An example of this is presented below.
Example: a 1 = 3, q = 2 - both parameters are greater than one.
Then the numerical sequence can be written like this:
3 6 12 24 48 ...
- If | q | less than one, that is, multiplying by it is equivalent to division, then a progression with similar conditions is a decreasing geometric progression. An example of this is presented below.
Example: a 1 = 6, q = 1/3 - a 1 is more than one, q is less.
Then the numerical sequence can be written as follows:
6 2 2/3 ... - any element is 3 times larger than the element following it.
- Alternating. If q <0, then the signs of the numbers in the sequence constantly alternate regardless of a 1 , and the elements neither increase nor decrease.
Example: a 1 = -3, q = -2 - both parameters are less than zero.
Then the numerical sequence can be written as follows:
-3, 6, -12, 24, ...
Formulas
For the convenient use of geometric progressions, there are many formulas:
- The formula of the z-th term. Allows you to calculate the element standing under a specific number without calculating the previous numbers.
Example: q = 3, a 1 = 4. It is required to calculate the fourth element of progression.
Solution: a 4 = 4 · 3 4-1 = 4 · 3 3 = 4 · 27 = 108.
- The sum of the first elements whose quantity is equal to z . Allows you to calculate the sum of all elements of the sequence up to a z inclusive.
Since (1- q ) is in the denominator, (1 - q) ≠ 0, therefore, q is not equal to 1.
Note: if q = 1, then the progression would be a series of infinitely repeating numbers.
The sum of the geometric progression, examples: a 1 = 2, q = -2. Count S 5 .
Solution: S 5 = 22 - calculation by the formula.
- Amount if | q | <1 and if z tends to infinity.
Example: a 1 = 2 , q = 0.5. Find the amount.
Solution: S z = 2 · = 4
If you calculate the sum of several members manually, you can see that it really tends to four.
S z = 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 3.9375 4
Some properties:
- Characteristic property. If the following condition holds for any z , then the given number series is a geometric progression:
a z 2 = a z -1 A z + 1
- Also, the square of any number of geometric progression is found by adding the squares of any two other numbers in the given row, if they are equidistant from this element.
a z 2 = a z - t 2 + a z + t 2 , where t is the distance between these numbers.
- Elements differ q times.
- The logarithms of the elements of progression also form a progression, but already arithmetic, that is, each of them is greater than the previous one by a certain number.
Examples of some classic problems
To better understand what geometric progression is, examples with a solution for grade 9 may help.
- Conditions: a 1 = 3, a 3 = 48. Find q .
Solution: each subsequent element is larger than the previous in q time. It is necessary to express some elements through others using a denominator.
Therefore, a 3 = q 2
With the substitution q = 4
- Conditions: a 2 = 6, a 3 = 12. Calculate S 6 .
Solution: To do this, just find q, the first element and substitute it in the formula.
a 3 = q · a 2 , therefore, q = 2
a 2 = q · A 1 , therefore a 1 = 3
S 6 = 189
- A 1 = 10, q = -2. Find the fourth element of progression.
Solution: for this it is enough to express the fourth element through the first and through the denominator.
a 4 = q 3 a 1 = -80
Application example:
- The bank's client made a deposit of 10,000 rubles, according to which each year the client will be added to the principal amount 6% of it. How much will be in the account after 4 years?
Solution: The initial amount is 10 thousand rubles. So, a year after the investment, the account will have an amount equal to 10000 + 10000 · 0.06 = 10000 · 1.06
Accordingly, the amount in the account in another year will be expressed as follows:
(10000 · 1.06) · 0.06 + 10000 · 1.06 = 1.06 · 1.06 · 10000
That is, every year the amount increases by 1.06 times. So, to find the amount of funds in the account after 4 years, it is enough to find the fourth element of the progression, which is given by the first element equal to 10 thousand, and the denominator equal to 1.06.
S = 1.06 · 1.06 · 1.06 · 1.06 · 10000 = 12625
Examples of tasks for calculating the amount:
Various tasks use geometric progression. An example for finding the amount can be set as follows:
a 1 = 4, q = 2, calculate S 5 .
Solution: all the data necessary for the calculation is known, you just need to substitute them in the formula.
S 5 = 124
- a 2 = 6, a 3 = 18. Calculate the sum of the first six elements.
Decision:
In the geome. each next element is q times larger than the previous one, that is, to calculate the sum, you need to know the element a 1 and the denominator q .
a 2 · q = a 3
q = 3
Similarly, we need to find a 1 , knowing a 2 and q .
a 1 · q = a 2
a 1 = 2
And then it’s enough to substitute the known data into the sum formula.
S 6 = 728.