Among all sequences of numbers, the geometric progression, which is considered in the course of class 9 algebra, is one of the most famous. What it is and how to solve a geometric progression - these questions are answered in this article.
A sequence of numbers that obeys the mathematical law
The name of this item is a general definition of geometric progression. The law by which it is described is quite simple: each next number differs from the previous one by a factor, which is called the βdenominatorβ. It can be denoted by the letter r. Then we can write the following equality:
a n + 1 = a n * r.
Here a n is a member of the progression with number n.
If r is greater than 1, then the progression will increase in absolute value (it may decrease if its first term has a negative sign). If r is less than unity, then the whole progression will tend to zero either from below (a 1 <0) or from above (a 1 > 0). In the case of a negative denominator (r <0), an alternating numerical sequence will take place (each positive term will be surrounded by two negative ones). Finally, if r is equal to unity, we get a simple set of numbers, which, as a rule, is not called progression.
An example of the type of progression under consideration is given below:
2, 3, 4, 5, 6, 75, ...
Here, the first term is 2, and the denominator is 1.5.
Important formulas
How to solve geometric progression in grade 9? For this, one should know not only its definition and understand what is at stake, but also remember two important formulas. The first of these is given below:
a n = a 1 * r n-1 .
The expression allows you to easily find an arbitrary element of the sequence, but for this you need to know two numbers: the denominator and the first element. To prove this formula is simple, you just need to remember the definition of geometric progression: the second element is obtained by multiplying the first by the denominator in the first degree, the third element by multiplying the first by the denominator in the second degree and so on. The usefulness of this expression is obvious: there is no need for a sequential restoration of the entire number series in order to find out what value its nth element will take.
The following formula is also useful in answering the question of how to solve geometric progression. We are talking about the sum of its elements, starting from the first and ending with the nth. The corresponding expression is given below:
S n = a 1 * (r n -1) / (r-1).
It is worth paying attention to its peculiarity: as in the formula for finding the nth element, here it is also sufficient to know the same two numbers (a 1 and r). This result is not surprising, because each member of the progression is associated with the marked numbers.
The following are a few examples that show how to solve a geometric progression.
Progression recovery
The first example of how to solve geometric progression has the following condition: it is known that two numbers 10 and 20 form the type of progression under consideration. Moreover, the numbers are the eighth and fifteenth elements of the series. It is necessary to restore the entire series, knowing that it should be decreasing.
This somewhat confusing condition of the problem should be examined carefully: since this is a decreasing series, the number 10 should be in the 15th position, and the 20th should be in 8. When proceeding to the solution, write down the corresponding equalities for each of the numbers:
a 8 = a 1 * r 7 and a 15 = a 1 * r 14 .
You have two equalities with two unknowns. Solve them by expressing from the first a 1 and substituting it into the second. It will turn out:
a 1 = a 8 * r -7 and a 15 = a 8 * r -7 * r 14 = a 8 * r 7 => r = 7 β (a 15 / a 8 ).
Now it remains to substitute the corresponding values ββfrom the condition and calculate the root of the seventh degree. It will turn out:
r = 7 β (a 15 / a 8 ) = 7 β (10/20) β 0.9057.
Substituting the resulting denominator into any of the expressions for the known nth element, we get a 1 :
a 1 = a 8 * r -7 = 20 * (0.9057) -7 β 40.0073.
Thus, you will find the first term and denominator, which means that you will restore the entire progression. First few members:
40.0073, 36.2346, 32.8177, 29.7230, ...
It is worth noting that when performing the calculations, rounding to 4 decimal places was used.
Finding an unknown member of a row
Now itβs worth considering another example: it is known that the seventh element of the series is 27, which is equal to the thirteenth term if the denominator is r = -2. How to solve geometric progression using these data? Very simple, you need to write out the formula for the 7th element:
a 7 = a 1 * r 6 .
Since in this equality only the number a 1 is unknown, express it:
a 1 = a 7 * r -6 .
Use the last equality by substituting it in the formula for the 13th term to be found. It will turn out:
a 13 = a 1 * r 12 = a 7 * r -6 * r 12 = a 7 * r 6 .
It remains to substitute the numbers and write down the answer:
a 13 = a 7 * r 6 = 27 * (- 2) 6 = 1728.
The resulting number shows how fast the geometric progression is growing.
Task for the amount
The last task, revealing the question of how to solve a geometric progression, is associated with finding the sum of several elements. Let a 1 = 1.5, r = 2. You should calculate the sum of the members of this series, starting from the 5th and ending with the 10th.
To get the answer to this question, apply the formula:
S 5 10 = S 10 - S 4 .
That is, first you need to find the sum of 10 elements, then the sum of the first 4 and subtract them from each other. Following the specified algorithm, it turns out:
S 10 = a 1 * (r n -1) / (r-1) = 1.5 * (2 10 -1) / (2-1) = 1534.5;
S 4 = a 1 * (r n -1) / (r-1) = 1.5 * (2 4 -1) / (2-1) = 22.5;
S 5 10 = 1534.5 - 22.5 = 1512.
It is worth noting that in the final formula the sum of exactly 4 terms was subtracted, since the fifth by the condition of the problem should participate in the sum.