Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
|
Definition |
Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference) |
geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression) |
Recurrent formula |
For any natural n |
For any natural n |
nth term formula |
a n = a 1 + d (n - 1) |
b n \u003d b 1 ∙ q n - 1, b n ≠ 0 |
| characteristic property |  |
|
| Sum of the first n terms |  |
 |
Examples of tasks with comments
Exercise 1
In arithmetic progression ( a n) a 1 = -6, a 2
According to the formula of the nth term:
a 22 = a 1+ d (22 - 1) = a 1+ 21d
By condition:
a 1= -6, so a 22= -6 + 21d.
It is necessary to find the difference of progressions:
d= a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = - 48.
Answer : a 22 = -48.
Task 2
Find the fifth term of the geometric progression: -3; 6;....
1st way (using n-term formula)
According to the formula of the n-th member of a geometric progression:
b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
Because b 1 = -3,
2nd way (using recursive formula)
Since the denominator of the progression is -2 (q = -2), then:
b 3 = 6 ∙ (-2) = -12;
b 4 = -12 ∙ (-2) = 24;
b 5 = 24 ∙ (-2) = -48.
Answer : b 5 = -48.
Task 3
In arithmetic progression ( a n) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.
For an arithmetic progression, the characteristic property has the form 
.
Therefore:
.
Substitute the data in the formula:
Answer: 95.
Task 4
In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.
To find the sum of the first n terms of an arithmetic progression, two formulas are used:
.
Which one in this case more convenient to use?
By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, and a 16 without finding d . Therefore, we use the first formula.
Answer: 368.
Task 5
In arithmetic progression a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.
According to the formula of the nth term:
a 22 = a 1 + d (22 – 1) = a 1+ 21d.
By condition, if a 1= -6, then a 22= -6 + 21d. It is necessary to find the difference of progressions:
d= a 2 – a 1 = -8 – (-6) = -2
a 22 = -6 + 21 ∙ (-2) = -48.
Answer : a 22 = -48.
Task 6
Several consecutive terms of a geometric progression are recorded:
Find the term of the progression, denoted by the letter x .
When solving, we use the formula for the nth term b n \u003d b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.
Substituting the found values into the formula, we get:
.
Answer : .
Task 7
From arithmetic progressions, given by the formula nth term, select the one for which the condition is met a 27 > 9:
Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:
.
Answer: 4.
Task 8
In arithmetic progression a 1= 3, d = -1.5. Specify highest value n , for which the inequality a n > -6.
Many have heard of an arithmetic progression, but not everyone is well aware of what it is. In this article, we will give the corresponding definition, and also consider the question of how to find the difference of an arithmetic progression, and give a number of examples.
Mathematical definition
So if we are talking about an arithmetic or algebraic progression (these concepts define the same thing), this means that there is some number series satisfying next law: every two adjacent numbers in the series differ by the same amount. Mathematically, this is written like this:
Here n means the number of the element a n in the sequence, and the number d is the difference of the progression (its name follows from the presented formula).
What does knowing the difference d mean? About how far apart adjacent numbers are. However, knowledge of d is necessary, but not sufficient condition to determine (restore) the entire progression. You need to know one more number, which can be absolutely any element of the series under consideration, for example, a 4, a10, but, as a rule, the first number is used, that is, a 1.
Formulas for determining the elements of the progression
In general, the information above is already enough to proceed to the solution specific tasks. Nevertheless, before an arithmetic progression is given, and it will be necessary to find its difference, we present a pair useful formulas, thereby facilitating the subsequent process of solving problems.
It is easy to show that any element of the sequence with number n can be found as follows:
a n \u003d a 1 + (n - 1) * d
Indeed, everyone can check this formula by simple enumeration: if we substitute n = 1, then we get the first element, if we substitute n = 2, then the expression gives the sum of the first number and the difference, and so on.
The conditions of many problems are compiled in such a way that for a known pair of numbers, the numbers of which are also given in the sequence, it is necessary to restore the entire number series (find the difference and the first element). Now we will solve this problem in a general way.
So, let's say we are given two elements with numbers n and m. Using the formula obtained above, we can compose a system of two equations:
a n \u003d a 1 + (n - 1) * d;
a m = a 1 + (m - 1) * d
For finding unknown quantities let's use the well-known simple trick solutions of such a system: we subtract pairwise the left and right parts, while the equality remains valid. We have:
a n \u003d a 1 + (n - 1) * d;
a n - a m = (n - 1) * d - (m - 1) * d = d * (n - m)
Thus, we have eliminated one unknown (a 1). Now we can write the final expression for determining d:
d = (a n - a m) / (n - m), where n > m
We have received very a simple formula: to calculate the difference d in accordance with the conditions of the problem, it is only necessary to take the ratio of the differences of the elements themselves and their serial numbers. Should focus on one important point attention: the differences are taken between the "senior" and "junior" members, that is, n > m ("senior" - meaning standing farther from the beginning of the sequence, its absolute value can be either greater or less than the "younger" element).
The expression for the difference d of the progression should be substituted into any of the equations at the beginning of the solution of the problem in order to obtain the value of the first term.
In our age of development computer technology many schoolchildren try to find solutions for their tasks on the Internet, so questions of this type often arise: find the difference of an arithmetic progression online. Upon such a request, the search engine will display a number of web pages, by going to which, you will need to enter the data known from the condition (it can be either two members of the progression, or the sum of some of them) and instantly get an answer. Nevertheless, such an approach to solving the problem is unproductive in terms of the development of the student and understanding the essence of the task assigned to him.
Solution without using formulas
Let's solve the first problem, while we will not use any of the above formulas. Let the elements of the series be given: a6 = 3, a9 = 18. Find the difference of the arithmetic progression.
Known elements are close to each other in a row. How many times must the difference d be added to the smallest one to get the largest one? Three times (the first time adding d, we get the 7th element, the second time - the eighth, finally, the third time - the ninth). What number must be added to three three times to get 18? This is the number five. Really:
Thus, the unknown difference is d = 5.
Of course, the solution could be done using the appropriate formula, but this was not done intentionally. Detailed explanation problem solving should be clear and a prime example, what arithmetic progression.
A task similar to the previous one
Now let's decide similar task, but change the input data. So, you should find if a3 = 2, a9 = 19.
Of course, you can resort again to the method of solving "on the forehead". But since the elements of the series are given, which are relatively far apart, such a method becomes not very convenient. But using the resulting formula will quickly lead us to the answer:
d \u003d (a 9 - a 3) / (9 - 3) \u003d (19 - 2) / (6) \u003d 17 / 6 ≈ 2.83
Here we have rounded the final number. How much this rounding led to an error can be judged by checking the result:
a 9 \u003d a 3 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 + 2.83 \u003d 18.98
This result differs by only 0.1% from the value given in the condition. Therefore, the used rounding to hundredths can be considered successful choice.
Tasks for applying the formula for an member
Consider classic example tasks to determine the unknown d: find the difference of the arithmetic progression if a1 = 12, a5 = 40.
When two unknown numbers are given algebraic sequence, and one of them is the element a 1 , then you do not need to think long, but you should immediately apply the formula for the a n member. In this case we have:
a 5 = a 1 + d * (5 - 1) => d = (a 5 - a 1) / 4 = (40 - 12) / 4 = 7
We got exact number when dividing, so it makes no sense to check the accuracy of the calculated result, as was done in the previous paragraph.
Let's solve another similar problem: we should find the difference of the arithmetic progression if a1 = 16, a8 = 37.
We use a similar approach to the previous one and get:
a 8 = a 1 + d * (8 - 1) => d = (a 8 - a 1) / 7 = (37 - 16) / 7 = 3
What else you should know about arithmetic progression
In addition to the task of finding unknown difference or individual elements, it is often necessary to solve problems of the sum of the first terms of a sequence. Consideration of these problems is beyond the scope of the topic of the article; nevertheless, for completeness of information, we present general formula for the sum of n numbers of the series:
∑ n i = 1 (a i) = n * (a 1 + a n) / 2
If every natural number n put in line real number a n , then they say that given number sequence :
a 1 , a 2 , a 3 , . . . , a n , . . . .
So, numerical sequence is a function of natural argument.
Number a 1 called the first member of the sequence , number a 2 — the second member of the sequence , number a 3 — third and so on. Number a n called nth member sequences , and the natural number n — his number .
From two neighboring members a n and a n +1 member sequences a n +1 called subsequent (towards a n ), a a n — previous (towards a n +1 ).
To specify a sequence, you must specify a method that allows you to find a sequence member with any number.
Often the sequence is given with nth term formulas , that is, a formula that allows you to determine a sequence member by its number.
For example,
sequence of positive odd numbers can be given by the formula
a n= 2n- 1,
and the sequence of alternating 1 and -1 - formula
b n = (-1)n +1 . ◄
The sequence can be determined recurrent formula, that is, a formula that expresses any member of the sequence, starting with some, through the previous (one or more) members.
For example,
if a 1 = 1 , a a n +1 = a n + 5
a 1 = 1,
a 2 = a 1 + 5 = 1 + 5 = 6,
a 3 = a 2 + 5 = 6 + 5 = 11,
a 4 = a 3 + 5 = 11 + 5 = 16,
a 5 = a 4 + 5 = 16 + 5 = 21.
If a a 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven members of the numerical sequence are set as follows:
a 1 = 1,
a 2 = 1,
a 3 = a 1 + a 2 = 1 + 1 = 2,
a 4 = a 2 + a 3 = 1 + 2 = 3,
a 5 = a 3 + a 4 = 2 + 3 = 5,
a 6 = a 4 + a 5 = 3 + 5 = 8,
a 7 = a 5 + a 6 = 5 + 8 = 13. ◄
Sequences can be final and endless .
The sequence is called ultimate if it has a finite number of members. The sequence is called endless if it has infinitely many members.
For example,
sequence of two-digit natural numbers:
10, 11, 12, 13, . . . , 98, 99
final.
Prime number sequence:
2, 3, 5, 7, 11, 13, . . .
endless. ◄
The sequence is called increasing , if each of its members, starting from the second, is greater than the previous one.
The sequence is called waning , if each of its members, starting from the second, is less than the previous one.
For example,
2, 4, 6, 8, . . . , 2n, . . . is an ascending sequence;
1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . is a descending sequence. ◄
A sequence whose elements do not decrease with increasing number, or, conversely, do not increase, is called monotonous sequence .
Monotonic sequences, in particular, are increasing sequences and decreasing sequences.
Arithmetic progression
Arithmetic progression a sequence is called, each member of which, starting from the second, is equal to the previous one, to which the same number is added.
a 1 , a 2 , a 3 , . . . , a n, . . .
is an arithmetic progression if for any natural number n condition is met:
a n +1 = a n + d,
where d - some number.
Thus, the difference between the next and the previous members of a given arithmetic progression is always constant:
a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.
Number d called the difference of an arithmetic progression.
To set an arithmetic progression, it is enough to specify its first term and difference.
For example,
if a 1 = 3, d = 4 , then the first five terms of the sequence are found as follows:
a 1 =3,
a 2 = a 1 + d = 3 + 4 = 7,
a 3 = a 2 + d= 7 + 4 = 11,
a 4 = a 3 + d= 11 + 4 = 15,
a 5 = a 4 + d= 15 + 4 = 19. ◄
For an arithmetic progression with the first term a 1 and difference d her n
a n = a 1 + (n- 1)d.
For example,
find the thirtieth term of an arithmetic progression
1, 4, 7, 10, . . .
a 1 =1, d = 3,
a 30 = a 1 + (30 - 1)d= 1 + 29· 3 = 88. ◄
a n-1 = a 1 + (n- 2)d,
a n= a 1 + (n- 1)d,
a n +1 = a 1 + nd,
then obviously
| a n=
| a n-1 + a n+1
|
| 2
|
each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the previous and subsequent members.
numbers a, b and c are consecutive members of some arithmetic progression if and only if one of them is equal to the arithmetic mean of the other two.
For example,
a n = 2n- 7 , is an arithmetic progression.
Let's use the statement above. We have:
a n = 2n- 7,
a n-1 = 2(n- 1) - 7 = 2n- 9,
a n+1 = 2(n+ 1) - 7 = 2n- 5.
Consequently,
| a n+1 + a n-1
| =
| 2n- 5 + 2n- 9
| = 2n- 7 = a n,
|
| 2
| 2
|
◄
Note that n -th member of an arithmetic progression can be found not only through a 1 , but also any previous a k
a n = a k + (n- k)d.
For example,
for a 5 can be written
a 5 = a 1 + 4d,
a 5 = a 2 + 3d,
a 5 = a 3 + 2d,
a 5 = a 4 + d. ◄
a n = a n-k + kd,
a n = a n+k - kd,
then obviously
| a n=
| a n-k
+ a n+k
|
| 2
|
any member of an arithmetic progression, starting from the second, is equal to half the sum of the members of this arithmetic progression equally spaced from it.
In addition, for any arithmetic progression, the equality is true:
a m + a n = a k + a l,
m + n = k + l.
For example,
in arithmetic progression
1) a 10 = 28 = (25 + 31)/2 = (a 9 + a 11 )/2;
2) 28 = a 10 = a 3 + 7d= 7 + 7 3 = 7 + 21 = 28;
3) a 10= 28 = (19 + 37)/2 = (a 7 + a 13)/2;
4) a 2 + a 12 = a 5 + a 9, because
a 2 + a 12= 4 + 34 = 38,
a 5 + a 9 = 13 + 25 = 38. ◄
S n= a 1 + a 2 + a 3 + . . .+ a n,
first n members of an arithmetic progression is equal to the product of half the sum of the extreme terms by the number of terms:
From this, in particular, it follows that if it is necessary to sum the terms
a k, a k +1 , . . . , a n,
then the previous formula retains its structure:
For example,
in arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, . . .
S 10 = 1 + 4 + . . . + 28 = (1 + 28) · 10/2 = 145;
10 + 13 + 16 + 19 + 22 + 25 + 28 = S 10 - S 3 = (10 + 28 ) · (10 - 4 + 1)/2 = 133. ◄
If an arithmetic progression is given, then the quantities a 1 , a n, d, n andS n linked by two formulas:
Therefore, if three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.
An arithmetic progression is a monotonic sequence. Wherein:
- if d > 0 , then it is increasing;
- if d < 0 , then it is decreasing;
- if d = 0 , then the sequence will be stationary.
Geometric progression
geometric progression a sequence is called, each term of which, starting from the second, is equal to the previous one, multiplied by the same number.
b 1 , b 2 , b 3 , . . . , b n, . . .
is a geometric progression if for any natural number n condition is met:
b n +1 = b n · q,
where q ≠ 0 - some number.
Thus, the ratio of the next term of this geometric progression to the previous one is a constant number:
b 2 / b 1 = b 3 / b 2 = . . . = b n +1 / b n = q.
Number q called denominator of a geometric progression.
To set a geometric progression, it is enough to specify its first term and denominator.
For example,
if b 1 = 1, q = -3 , then the first five terms of the sequence are found as follows:
b 1 = 1,
b 2 = b 1 · q = 1 · (-3) = -3,
b 3 = b 2 · q= -3 · (-3) = 9,
b 4 = b 3 · q= 9 · (-3) = -27,
b 5 = b 4 · q= -27 · (-3) = 81. ◄
b 1 and denominator q her n -th term can be found by the formula:
b n = b 1 · q n -1 .
For example,
find the seventh term of a geometric progression 1, 2, 4, . . .
b 1 = 1, q = 2,
b 7 = b 1 · q 6 = 1 2 6 = 64. ◄
bn-1 = b 1 · q n -2 ,
b n = b 1 · q n -1 ,
b n +1 = b 1 · q n,
then obviously
b n 2 = b n -1 · b n +1 ,
each member of the geometric progression, starting from the second, is equal to the geometric mean (proportional) of the previous and subsequent members.
Since the converse is also true, the following assertion holds:
numbers a, b and c are consecutive members of some geometric progression if and only if the square of one of them is equal to the product the other two, that is, one of the numbers is the geometric mean of the other two.
For example,
let us prove that the sequence given by the formula b n= -3 2 n , is a geometric progression. Let's use the statement above. We have:
b n= -3 2 n,
b n -1 = -3 2 n -1 ,
b n +1 = -3 2 n +1 .
Consequently,
b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) (-3 2 n +1 ) = b n -1 · b n +1 ,
which proves the required assertion. ◄
Note that n th term of a geometric progression can be found not only through b 1 , but also any previous term b k , for which it suffices to use the formula
b n = b k · q n - k.
For example,
for b 5 can be written
b 5 = b 1 · q 4 ,
b 5 = b 2 · q 3,
b 5 = b 3 · q2,
b 5 = b 4 · q. ◄
b n = b k · q n - k,
b n = b n - k · q k,
then obviously
b n 2 = b n - k· b n + k
the square of any member of a geometric progression, starting from the second, is equal to the product of the members of this progression equidistant from it.
In addition, for any geometric progression, the equality is true:
b m· b n= b k· b l,
m+ n= k+ l.
For example,
exponentially
1) b 6 2 = 32 2 = 1024 = 16 · 64 = b 5 · b 7 ;
2) 1024 = b 11 = b 6 · q 5 = 32 · 2 5 = 1024;
3) b 6 2 = 32 2 = 1024 = 8 · 128 = b 4 · b 8 ;
4) b 2 · b 7 = b 4 · b 5 , because
b 2 · b 7 = 2 · 64 = 128,
b 4 · b 5 = 8 · 16 = 128. ◄
S n= b 1 + b 2 + b 3 + . . . + b n
first n members of a geometric progression with a denominator q ≠ 0 calculated by the formula:
And when q = 1 - according to the formula
S n= n.b. 1
Note that if we need to sum the terms
b k, b k +1 , . . . , b n,
then the formula is used:
| S n- S k -1 = b k + b k +1 + . . . + b n = b k · | 1 - q n -
k +1
| . |
| 1 - q
|
For example,
exponentially 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .
S 10 = 1 + 2 + . . . + 512 = 1 · (1 - 2 10) / (1 - 2) = 1023;
64 + 128 + 256 + 512 = S 10 - S 6 = 64 · (1 - 2 10-7+1) / (1 - 2) = 960. ◄
If given geometric progression, then the quantities b 1 , b n, q, n and S n linked by two formulas:
Therefore, if the values of any three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.
For a geometric progression with the first term b 1 and denominator q the following take place monotonicity properties :
- the progression is increasing if one of the following conditions is met:
b 1 > 0 and q> 1;
b 1 < 0 and 0 < q< 1;
- A progression is decreasing if one of the following conditions is met:
b 1 > 0 and 0 < q< 1;
b 1 < 0 and q> 1.
If a q< 0 , then the geometric progression is sign-alternating: its odd-numbered terms have the same sign as its first term, and even-numbered terms have the opposite sign. It is clear that an alternating geometric progression is not monotonic.
Product of the first n terms of a geometric progression can be calculated by the formula:
P n= b 1 · b 2 · b 3 · . . . · b n = (b 1 · b n) n / 2 .
For example,
1 · 2 · 4 · 8 · 16 · 32 · 64 · 128 = (1 · 128) 8/2 = 128 4 = 268 435 456;
3 · 6 · 12 · 24 · 48 = (3 · 48) 5/2 = (144 1/2) 5 = 12 5 = 248 832.◄
Infinitely decreasing geometric progression
Infinitely decreasing geometric progression is called an infinite geometric progression whose denominator modulus is less than 1 , that is
|q| < 1 .
Note that an infinitely decreasing geometric progression may not be a decreasing sequence. This fits the case
1 < q< 0 .
With such a denominator, the sequence is sign-alternating. For example,
1, - 1 / 2 , 1 / 4 , - 1 / 8 , . . . .
The sum of an infinitely decreasing geometric progression name the number to which the sum of the first n terms of the progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula
| S= b 1 + b 2 + b 3 + . . . = | b 1
| . |
| 1 - q
|
For example,
10 + 1 + 0,1 + 0,01 + . . . = 10 / (1 - 0,1) = 11 1 / 9 ,
10 - 1 + 0,1 - 0,01 + . . . = 10 / (1 + 0,1) = 9 1 / 11 . ◄
Relationship between arithmetic and geometric progressions
Arithmetic and geometric progressions are closely related. Let's consider just two examples.
a 1 , a 2 , a 3 , . . . d , then
b a 1 , b a 2 , b a 3 , . . . b d .
For example,
1, 3, 5, . . . — arithmetic progression with difference 2 and
7 1 , 7 3 , 7 5 , . . . is a geometric progression with a denominator 7 2 . ◄
b 1 , b 2 , b 3 , . . . is a geometric progression with a denominator q , then
log a b 1, log a b 2, log a b 3, . . . — arithmetic progression with difference log aq .
For example,
2, 12, 72, . . . is a geometric progression with a denominator 6 and
lg 2, lg 12, lg 72, . . . — arithmetic progression with difference lg 6 . ◄
Yes, yes: arithmetic progression is not a toy for you :) Well, friends, if you are reading this text, then the internal cap evidence tells me that you still do not know what an arithmetic progression is, but you really (no, like this: SOOOOO!) want to know. Therefore, I will not torment you with long introductions and will immediately get down to business.
To start, a couple of examples. Consider several sets of numbers:
- 1; 2; 3; 4; ...
- 15; 20; 25; 30; ...
- $\sqrt(2);\ 2\sqrt(2);\ 3\sqrt(2);...$
What do all these sets have in common? At first glance, nothing. But actually there is something. Namely: each next element differs from the previous one by the same number.
Judge for yourself. The first set is just consecutive numbers, each one more than the previous one. In the second case, the difference between standing numbers is already equal to five, but this difference is still constant. In the third case, there are roots in general. However, $2\sqrt(2)=\sqrt(2)+\sqrt(2)$, while $3\sqrt(2)=2\sqrt(2)+\sqrt(2)$, i.e. in which case each next element simply increases by $\sqrt(2)$ (and don't be scared that this number is irrational).
So: all such sequences are just called arithmetic progressions. Let's give a strict definition:
Definition. A sequence of numbers in which each next differs from the previous one by exactly the same amount is called an arithmetic progression. The very amount by which the numbers differ is called the progression difference and is most often denoted by the letter $d$.
Notation: $\left(((a)_(n)) \right)$ is the progression itself, $d$ is its difference.
And just a couple important notes. First, progression is considered only orderly sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. You can't rearrange or swap numbers.
Secondly, the sequence itself can be either finite or infinite. For example, the set (1; 2; 3) is obviously a finite arithmetic progression. But if you write something in the spirit (1; 2; 3; 4; ...) - this is already infinite progression. The ellipsis after the four, as it were, hints that quite a lot of numbers go further. Infinitely many, for example. :)
I would also like to note that progressions are increasing and decreasing. We have already seen increasing ones - the same set (1; 2; 3; 4; ...). Here are examples of decreasing progressions:
- 49; 41; 33; 25; 17; ...
- 17,5; 12; 6,5; 1; −4,5; −10; ...
- $\sqrt(5);\ \sqrt(5)-1;\ \sqrt(5)-2;\ \sqrt(5)-3;...$
OK OK: last example may seem overly complicated. But the rest, I think, you understand. Therefore, we introduce new definitions:
Definition. An arithmetic progression is called:
- increasing if each next element is greater than the previous one;
- decreasing, if, on the contrary, each subsequent element is less than the previous one.
In addition, there are so-called "stationary" sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).
Only one question remains: how to distinguish an increasing progression from a decreasing one? Fortunately, everything here depends only on the sign of the number $d$, i.e. progression differences:
- If $d \gt 0$, then the progression is increasing;
- If $d \lt 0$, then the progression is obviously decreasing;
- Finally, there is the case $d=0$, in which case the entire progression reduces to the stationary sequence same numbers: (1; 1; 1; 1; ...) etc.
Let's try to calculate the difference $d$ for the three decreasing progressions above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract from the number on the right, the number on the left. It will look like this:
- 41−49=−8;
- 12−17,5=−5,5;
- $\sqrt(5)-1-\sqrt(5)=-1$.
As you can see, in all three cases the difference really turned out to be negative. And now that we have more or less figured out the definitions, it's time to figure out how progressions are described and what properties they have.
Members of the progression and the recurrent formula
Since the elements of our sequences cannot be interchanged, they can be numbered:
\[\left(((a)_(n)) \right)=\left\( ((a)_(1)),\ ((a)_(2)),((a)_(3 )),... \right\)\]
Individual elements of this set are called members of the progression. They are indicated in this way with the help of a number: the first member, the second member, and so on.
In addition, as we already know, neighboring members of the progression are related by the formula:
\[((a)_(n))-((a)_(n-1))=d\Rightarrow ((a)_(n))=((a)_(n-1))+d \]
In short, to find the $n$th term of the progression, you need to know the $n-1$th term and the difference $d$. Such a formula is called recurrent, because with its help you can find any number, only knowing the previous one (and in fact, all the previous ones). This is very inconvenient, so there is a more tricky formula that reduces any calculation to the first term and the difference:
\[((a)_(n))=((a)_(1))+\left(n-1 \right)d\]
You have probably come across this formula before. They like to give it in all sorts of reference books and reshebniks. And in any sensible textbook on mathematics, it is one of the first.
However, I suggest you practice a little.
Task number 1. Write down the first three terms of the arithmetic progression $\left(((a)_(n)) \right)$ if $((a)_(1))=8,d=-5$.
Solution. So, we know the first term $((a)_(1))=8$ and the progression difference $d=-5$. Let's use the formula just given and substitute $n=1$, $n=2$ and $n=3$:
\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)d; \\ & ((a)_(1))=((a)_(1))+\left(1-1 \right)d=((a)_(1))=8; \\ & ((a)_(2))=((a)_(1))+\left(2-1 \right)d=((a)_(1))+d=8-5= 3; \\ & ((a)_(3))=((a)_(1))+\left(3-1 \right)d=((a)_(1))+2d=8-10= -2. \\ \end(align)\]
Answer: (8; 3; -2)
That's all! Note that our progression is decreasing.
Of course, $n=1$ could not have been substituted - we already know the first term. However, by substituting the unit, we made sure that even for the first term our formula works. In other cases, everything came down to banal arithmetic.
Task number 2. Write out the first three terms of an arithmetic progression if its seventh term is −40 and its seventeenth term is −50.
Solution. We write the condition of the problem in the usual terms:
\[((a)_(7))=-40;\quad ((a)_(17))=-50.\]
\[\left\( \begin(align) & ((a)_(7))=((a)_(1))+6d \\ & ((a)_(17))=((a) _(1))+16d \\ \end(align) \right.\]
\[\left\( \begin(align) & ((a)_(1))+6d=-40 \\ & ((a)_(1))+16d=-50 \\ \end(align) \right.\]
I put the sign of the system because these requirements must be met simultaneously. And now we note that if we subtract the first equation from the second equation (we have the right to do this, because we have a system), we get this:
\[\begin(align) & ((a)_(1))+16d-\left(((a)_(1))+6d \right)=-50-\left(-40 \right); \\ & ((a)_(1))+16d-((a)_(1))-6d=-50+40; \\ & 10d=-10; \\&d=-1. \\ \end(align)\]
Just like that, we found the progression difference! It remains to substitute the found number in any of the equations of the system. For example, in the first:
\[\begin(matrix) ((a)_(1))+6d=-40;\quad d=-1 \\ \Downarrow \\ ((a)_(1))-6=-40; \\ ((a)_(1))=-40+6=-34. \\ \end(matrix)\]
Now, knowing the first term and the difference, it remains to find the second and third terms:
\[\begin(align) & ((a)_(2))=((a)_(1))+d=-34-1=-35; \\ & ((a)_(3))=((a)_(1))+2d=-34-2=-36. \\ \end(align)\]
Ready! Problem solved.
Answer: (-34; -35; -36)
pay attention to curious property progression that we discovered: if we take the $n$th and $m$th terms and subtract them from each other, then we get the difference of the progression multiplied by the number $n-m$:
\[((a)_(n))-((a)_(m))=d\cdot \left(n-m \right)\]
Simple but very useful property, which you definitely need to know - with its help you can significantly speed up the solution of many problems in progressions. Here bright to that example:
Task number 3. The fifth term of the arithmetic progression is 8.4, and its tenth term is 14.4. Find the fifteenth term of this progression.
Solution. Since $((a)_(5))=8.4$, $((a)_(10))=14.4$, and we need to find $((a)_(15))$, we note following:
\[\begin(align) & ((a)_(15))-((a)_(10))=5d; \\ & ((a)_(10))-((a)_(5))=5d. \\ \end(align)\]
But by condition $((a)_(10))-((a)_(5))=14.4-8.4=6$, so $5d=6$, whence we have:
\[\begin(align) & ((a)_(15))-14,4=6; \\ & ((a)_(15))=6+14.4=20.4. \\ \end(align)\]
Answer: 20.4
That's all! We did not need to compose any systems of equations and calculate the first term and the difference - everything was decided in just a couple of lines.
Now let's consider another type of problem - the search for negative and positive members of the progression. It is no secret that if the progression increases, while its first term is negative, then sooner or later positive terms will appear in it. And vice versa: the terms of a decreasing progression will sooner or later become negative.
At the same time, it is far from always possible to find this moment “on the forehead”, sequentially sorting through the elements. Often, problems are designed in such a way that without knowing the formulas, calculations would take several sheets - we would just fall asleep until we found the answer. Therefore, we will try to solve these problems in a faster way.
Task number 4. How many negative terms in an arithmetic progression -38.5; -35.8; …?
Solution. So, $((a)_(1))=-38.5$, $((a)_(2))=-35.8$, from which we immediately find the difference:
Note that the difference is positive, so the progression is increasing. The first term is negative, so indeed at some point we will stumble upon positive numbers. The only question is when this will happen.
Let's try to find out: how long (i.e., up to what natural number $n$) the negativity of the terms is preserved:
\[\begin(align) & ((a)_(n)) \lt 0\Rightarrow ((a)_(1))+\left(n-1 \right)d \lt 0; \\ & -38.5+\left(n-1 \right)\cdot 2.7 \lt 0;\quad \left| \cdot 10 \right. \\ & -385+27\cdot \left(n-1 \right) \lt 0; \\ & -385+27n-27 \lt 0; \\ & 27n \lt 412; \\ & n \lt 15\frac(7)(27)\Rightarrow ((n)_(\max ))=15. \\ \end(align)\]
The last line needs clarification. So we know that $n \lt 15\frac(7)(27)$. On the other hand, only integer values of the number will suit us (moreover: $n\in \mathbb(N)$), so the largest allowable number is precisely $n=15$, and in no case 16.
Task number 5. In arithmetic progression $(()_(5))=-150,(()_(6))=-147$. Find the number of the first positive term of this progression.
This would be exactly the same problem as the previous one, but we don't know $((a)_(1))$. But the neighboring terms are known: $((a)_(5))$ and $((a)_(6))$, so we can easily find the progression difference:
In addition, let's try to express the fifth term in terms of the first and the difference using the standard formula:
\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)\cdot d; \\ & ((a)_(5))=((a)_(1))+4d; \\ & -150=((a)_(1))+4\cdot 3; \\ & ((a)_(1))=-150-12=-162. \\ \end(align)\]
Now we proceed by analogy with the previous problem. We find out at what point in our sequence positive numbers will appear:
\[\begin(align) & ((a)_(n))=-162+\left(n-1 \right)\cdot 3 \gt 0; \\ & -162+3n-3 \gt 0; \\ & 3n \gt 165; \\ & n \gt 55\Rightarrow ((n)_(\min ))=56. \\ \end(align)\]
Minimum integer solution given inequality is the number 56.
Please note: in last assignment everything came down to strict inequality, so the option $n=55$ will not suit us.
Now that we have learned how to solve simple problems, let's move on to more complex ones. But first, let's learn another very useful property of arithmetic progressions, which will save us a lot of time and unequal cells in the future. :)
Arithmetic mean and equal indents
Consider several consecutive terms of the increasing arithmetic progression $\left(((a)_(n)) \right)$. Let's try to mark them on a number line:
Arithmetic progression members on the number lineI specifically noted the arbitrary members $((a)_(n-3)),...,((a)_(n+3))$, and not any $((a)_(1)) ,\ ((a)_(2)),\ ((a)_(3))$ etc. Because the rule, which I will now tell you, works the same for any "segments".
And the rule is very simple. Let's remember the recursive formula and write it down for all marked members:
\[\begin(align) & ((a)_(n-2))=((a)_(n-3))+d; \\ & ((a)_(n-1))=((a)_(n-2))+d; \\ & ((a)_(n))=((a)_(n-1))+d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n+1))+d; \\ \end(align)\]
However, these equalities can be rewritten differently:
\[\begin(align) & ((a)_(n-1))=((a)_(n))-d; \\ & ((a)_(n-2))=((a)_(n))-2d; \\ & ((a)_(n-3))=((a)_(n))-3d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(n+3))=((a)_(n))+3d; \\ \end(align)\]
Well, so what? But the fact that the terms $((a)_(n-1))$ and $((a)_(n+1))$ lie at the same distance from $((a)_(n)) $. And this distance is equal to $d$. The same can be said about the terms $((a)_(n-2))$ and $((a)_(n+2))$ - they are also removed from $((a)_(n))$ by the same distance equal to $2d$. You can continue indefinitely, but the picture illustrates the meaning well
The members of the progression lie at the same distance from the center What does this mean for us? This means that you can find $((a)_(n))$ if the neighboring numbers are known:
\[((a)_(n))=\frac(((a)_(n-1))+((a)_(n+1)))(2)\]
We have deduced a magnificent statement: each member of an arithmetic progression is equal to the arithmetic mean of the neighboring members! Moreover, we can deviate from our $((a)_(n))$ to the left and to the right not by one step, but by $k$ steps — and still the formula will be correct:
\[((a)_(n))=\frac(((a)_(n-k))+((a)_(n+k)))(2)\]
Those. we can easily find some $((a)_(150))$ if we know $((a)_(100))$ and $((a)_(200))$, because $(( a)_(150))=\frac(((a)_(100))+((a)_(200)))(2)$. At first glance, it may seem that this fact does not give us anything useful. However, in practice, many tasks are specially "sharpened" for the use of the arithmetic mean. Take a look:
Task number 6. Find all values of $x$ such that the numbers $-6((x)^(2))$, $x+1$ and $14+4((x)^(2))$ are consecutive members of an arithmetic progression (in specified order).
Solution. Because the indicated numbers are members of the progression, they satisfy the arithmetic mean condition: the central element $x+1$ can be expressed in terms of neighboring elements:
\[\begin(align) & x+1=\frac(-6((x)^(2))+14+4((x)^(2)))(2); \\ & x+1=\frac(14-2((x)^(2)))(2); \\ & x+1=7-((x)^(2)); \\ & ((x)^(2))+x-6=0. \\ \end(align)\]
It turned out classic quadratic equation. Its roots: $x=2$ and $x=-3$ are the answers.
Answer: -3; 2.
Task number 7. Find the values of $$ such that the numbers $-1;4-3;(()^(2))+1$ form an arithmetic progression (in that order).
Solution. Let's express again middle member through the arithmetic mean of neighboring members:
\[\begin(align) & 4x-3=\frac(x-1+((x)^(2))+1)(2); \\ & 4x-3=\frac(((x)^(2))+x)(2);\quad \left| \cdot 2\right.; \\ & 8x-6=((x)^(2))+x; \\ & ((x)^(2))-7x+6=0. \\ \end(align)\]
Another quadratic equation. And again two roots: $x=6$ and $x=1$.
Answer: 1; 6.
If in the process of solving a problem you get some brutal numbers, or you are not completely sure of the correctness of the answers found, then there is a wonderful trick that allows you to check: did we solve the problem correctly?
Let's say in problem 6 we got answers -3 and 2. How can we check that these answers are correct? Let's just plug them into the original condition and see what happens. Let me remind you that we have three numbers ($-6(()^(2))$, $+1$ and $14+4(()^(2))$), which should form an arithmetic progression. Substitute $x=-3$:
\[\begin(align) & x=-3\Rightarrow \\ & -6((x)^(2))=-54; \\ &x+1=-2; \\ & 14+4((x)^(2))=50. \end(align)\]
We got the numbers -54; −2; 50 that differ by 52 is undoubtedly an arithmetic progression. The same thing happens for $x=2$:
\[\begin(align) & x=2\Rightarrow \\ & -6((x)^(2))=-24; \\ &x+1=3; \\ & 14+4((x)^(2))=30. \end(align)\]
Again a progression, but with a difference of 27. Thus, the problem is solved correctly. Those who wish can check the second task on their own, but I’ll say right away: everything is correct there too.
In general, while solving the last tasks, we stumbled upon another interesting fact, which also needs to be remembered:
If three numbers are such that the second is the average of the first and last, then these numbers form an arithmetic progression.
In the future, understanding this statement will allow us to literally “construct” the necessary progressions based on the condition of the problem. But before we engage in such a "construction", we should pay attention to one more fact, which directly follows from what has already been considered.
Grouping and sum of elements
Let's go back to numerical axis. We note there several members of the progression, between which, perhaps. worth a lot of other members:
6 elements marked on the number lineLet's try to express the "left tail" in terms of $((a)_(n))$ and $d$, and the "right tail" in terms of $((a)_(k))$ and $d$. It's very simple:
\[\begin(align) & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(k-1))=((a)_(k))-d; \\ & ((a)_(k-2))=((a)_(k))-2d. \\ \end(align)\]
Now note that the following sums are equal:
\[\begin(align) & ((a)_(n))+((a)_(k))=S; \\ & ((a)_(n+1))+((a)_(k-1))=((a)_(n))+d+((a)_(k))-d= S; \\ & ((a)_(n+2))+((a)_(k-2))=((a)_(n))+2d+((a)_(k))-2d= S. \end(align)\]
Simply put, if we consider as a start two elements of the progression, which in total equal to some number $S$, and then we start stepping from these elements to opposite sides(towards each other or vice versa to remove), then the sums of the elements that we will stumble upon will also be equal$S$. This can be best represented graphically:
Same indents give equal sums Understanding this fact will allow us to solve problems fundamentally more high level complexity than those discussed above. For example, these:
Task number 8. Determine the difference of an arithmetic progression in which the first term is 66, and the product of the second and twelfth terms is the smallest possible.
Solution. Let's write down everything we know:
\[\begin(align) & ((a)_(1))=66; \\&d=? \\ & ((a)_(2))\cdot ((a)_(12))=\min . \end(align)\]
So, we do not know the difference of the progression $d$. Actually, the whole solution will be built around the difference, since the product $((a)_(2))\cdot ((a)_(12))$ can be rewritten as follows:
\[\begin(align) & ((a)_(2))=((a)_(1))+d=66+d; \\ & ((a)_(12))=((a)_(1))+11d=66+11d; \\ & ((a)_(2))\cdot ((a)_(12))=\left(66+d \right)\cdot \left(66+11d \right)= \\ & =11 \cdot \left(d+66 \right)\cdot \left(d+6 \right). \end(align)\]
For those in the tank: I took out common factor 11 from the second bracket. Thus, the desired product is a quadratic function with respect to the variable $d$. Therefore, consider the function $f\left(d \right)=11\left(d+66 \right)\left(d+6 \right)$ - its graph will be a parabola with branches up, because if we open the brackets, we get:
\[\begin(align) & f\left(d \right)=11\left(((d)^(2))+66d+6d+66\cdot 6 \right)= \\ & =11(( d)^(2))+11\cdot 72d+11\cdot 66\cdot 6 \end(align)\]
As you can see, the coefficient at the highest term is 11 - this is positive number, so we are really dealing with a parabola with branches up:
schedule quadratic function- parabola
Note: minimum value this parabola takes $((d)_(0))$ at its vertex with abscissa. Of course, we can calculate this abscissa according to the standard scheme (there is a formula $((d)_(0))=(-b)/(2a)\;$), but it would be much more reasonable to note that the desired vertex lies on the axis symmetry of the parabola, so the point $((d)_(0))$ is equidistant from the roots of the equation $f\left(d \right)=0$:
\[\begin(align) & f\left(d\right)=0; \\ & 11\cdot \left(d+66 \right)\cdot \left(d+6 \right)=0; \\ & ((d)_(1))=-66;\quad ((d)_(2))=-6. \\ \end(align)\]
That is why I was in no hurry to open the brackets: in the original form, the roots were very, very easy to find. Therefore, the abscissa is equal to the mean arithmetic numbers-66 and -6:
\[((d)_(0))=\frac(-66-6)(2)=-36\]
What gives us the discovered number? With it, the required product takes smallest value(By the way, we did not calculate $((y)_(\min ))$ - we are not required to do this). At the same time, this number is the difference of the initial progression, i.e. we found the answer. :)
Answer: -36
Task number 9. Insert three numbers between the numbers $-\frac(1)(2)$ and $-\frac(1)(6)$ so that together with the given numbers they form an arithmetic progression.
Solution. In fact, we need to make a sequence of five numbers, with the first and last number already known. Denote the missing numbers by the variables $x$, $y$ and $z$:
\[\left(((a)_(n)) \right)=\left\( -\frac(1)(2);x;y;z;-\frac(1)(6) \right\ )\]
Note that the number $y$ is the "middle" of our sequence - it is equidistant from the numbers $x$ and $z$, and from the numbers $-\frac(1)(2)$ and $-\frac(1)( 6)$. And if from the numbers $x$ and $z$ we are in this moment we cannot get $y$, then the situation is different with the ends of the progression. Remember the arithmetic mean:
Now, knowing $y$, we will find the remaining numbers. Note that $x$ lies between $-\frac(1)(2)$ and $y=-\frac(1)(3)$ just found. That's why
Arguing similarly, we find the remaining number:
Ready! We found all three numbers. Let's write them down in the answer in the order in which they should be inserted between the original numbers.
Answer: $-\frac(5)(12);\ -\frac(1)(3);\ -\frac(1)(4)$
Task number 10. Between the numbers 2 and 42, insert several numbers that, together with the given numbers, form an arithmetic progression, if it is known that the sum of the first, second, and last of the inserted numbers is 56.
Solution. Even more difficult task, which, however, is solved in the same way as the previous ones - through the arithmetic mean. The problem is that we don't know exactly how many numbers to insert. Therefore, for definiteness, we assume that after inserting there will be exactly $n$ numbers, and the first of them is 2, and the last is 42. In this case, the desired arithmetic progression can be represented as:
\[\left(((a)_(n)) \right)=\left\( 2;((a)_(2));((a)_(3));...;(( a)_(n-1));42 \right\)\]
\[((a)_(2))+((a)_(3))+((a)_(n-1))=56\]
Note, however, that the numbers $((a)_(2))$ and $((a)_(n-1))$ are obtained from the numbers 2 and 42 standing at the edges by one step towards each other, i.e. . to the center of the sequence. And this means that
\[((a)_(2))+((a)_(n-1))=2+42=44\]
But then the above expression can be rewritten like this:
\[\begin(align) & ((a)_(2))+((a)_(3))+((a)_(n-1))=56; \\ & \left(((a)_(2))+((a)_(n-1)) \right)+((a)_(3))=56; \\ & 44+((a)_(3))=56; \\ & ((a)_(3))=56-44=12. \\ \end(align)\]
Knowing $((a)_(3))$ and $((a)_(1))$, we can easily find the progression difference:
\[\begin(align) & ((a)_(3))-((a)_(1))=12-2=10; \\ & ((a)_(3))-((a)_(1))=\left(3-1 \right)\cdot d=2d; \\ & 2d=10\Rightarrow d=5. \\ \end(align)\]
It remains only to find the remaining members:
\[\begin(align) & ((a)_(1))=2; \\ & ((a)_(2))=2+5=7; \\ & ((a)_(3))=12; \\ & ((a)_(4))=2+3\cdot 5=17; \\ & ((a)_(5))=2+4\cdot 5=22; \\ & ((a)_(6))=2+5\cdot 5=27; \\ & ((a)_(7))=2+6\cdot 5=32; \\ & ((a)_(8))=2+7\cdot 5=37; \\ & ((a)_(9))=2+8\cdot 5=42; \\ \end(align)\]
Thus, already at the 9th step we will come to the left end of the sequence - the number 42. In total, only 7 numbers had to be inserted: 7; 12; 17; 22; 27; 32; 37.
Answer: 7; 12; 17; 22; 27; 32; 37
Text tasks with progressions
In conclusion, I would like to consider a couple of simple tasks. Well, as simple ones: for most students who study mathematics at school and have not read what is written above, these tasks may seem like a gesture. Nevertheless, it is precisely such tasks that come across in the OGE and the USE in mathematics, so I recommend that you familiarize yourself with them.
Task number 11. The team produced 62 parts in January, and in each next month produced 14 parts more than in the previous one. How many parts did the brigade produce in November?
Solution. Obviously, the number of parts, painted by month, will be an increasing arithmetic progression. And:
\[\begin(align) & ((a)_(1))=62;\quad d=14; \\ & ((a)_(n))=62+\left(n-1 \right)\cdot 14. \\ \end(align)\]
November is the 11th month of the year, so we need to find $((a)_(11))$:
\[((a)_(11))=62+10\cdot 14=202\]
Therefore, 202 parts will be manufactured in November.
Task number 12. The bookbinding workshop bound 216 books in January, and each month it bound 4 more books than the previous month. How many books did the workshop bind in December?
Solution. All the same:
$\begin(align) & ((a)_(1))=216;\quad d=4; \\ & ((a)_(n))=216+\left(n-1 \right)\cdot 4. \\ \end(align)$
December is the last, 12th month of the year, so we are looking for $((a)_(12))$:
\[((a)_(12))=216+11\cdot 4=260\]
This is the answer - 260 books will be bound in December.
Well, if you have read this far, I hasten to congratulate you: “course young fighter» by arithmetic progressions you have successfully passed. You can safely move on to the next lesson, where we will study the progression sum formula, as well as important and very useful consequences from her.
Someone treats the word "progression" with caution, as a very complex term from sections higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi counter (where they still remain). And get the gist (and there is nothing more important in mathematics than "getting the gist") arithmetic sequence It's not that hard once you understand a few basic concepts.
Mathematical number sequence
It is customary to call a numerical sequence a series of numbers, each of which has its own number.
and 1 is the first member of the sequence;
and 2 is the second member of the sequence;
and 7 is the seventh member of the sequence;
and n is the nth member of the sequence;
However, not any arbitrary set of figures and numbers interests us. We will focus our attention on a numerical sequence in which the value of the n-th member is related to its ordinal number by a dependence that can be clearly formulated mathematically. In other words: numerical value The nth number is some function of n.
a - value of a member of the numerical sequence;
n - his serial number;
f(n) is a function where the ordinal in the numeric sequence n is the argument.
Definition
An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth member of an arithmetic sequence is as follows:
a n - the value of the current member of the arithmetic progression;
a n+1 - the formula of the next number;
d - difference (a certain number).
It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one, and such an arithmetic progression will be increasing.
In the graph below, it is easy to see why the number sequence is called "increasing".
In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.
The value of the specified member
Sometimes it is necessary to determine the value of some arbitrary term a n of an arithmetic progression. You can do this by calculating successively the values of all members of the arithmetic progression, from the first to the desired one. However, this way is not always acceptable if, for example, it is necessary to find the value of the five thousandth or eight millionth term. The traditional calculation will take a long time. However, a specific arithmetic progression can be investigated using certain formulas. There is also a formula for the nth term: the value of any member of an arithmetic progression can be determined as the sum of the first member of the progression with the difference of the progression, multiplied by the number of the desired member, minus one.
The formula is universal for increasing and decreasing progression.
An example of calculating the value of a given member
Let's solve the following problem of finding the value of the n-th member of an arithmetic progression.
Condition: there is an arithmetic progression with parameters:
The first member of the sequence is 3;
The difference in the number series is 1.2.
Task: it is necessary to find the value of 214 terms
Solution: to determine the value of a given member, we use the formula:
a(n) = a1 + d(n-1)
Substituting the data from the problem statement into the expression, we have:
a(214) = a1 + d(n-1)
a(214) = 3 + 1.2 (214-1) = 258.6
Answer: The 214th member of the sequence is equal to 258.6.
The advantages of this calculation method are obvious - the entire solution takes no more than 2 lines.
Sum of a given number of members
Very often, in a given arithmetic series, it is required to determine the sum of the values of some of its segments. It also doesn't need to calculate the values of each term and then sum them up. This method is applicable if the number of terms whose sum must be found is small. In other cases, it is more convenient to use the following formula.
The sum of the members of an arithmetic progression from 1 to n is equal to the sum of the first and nth members, multiplied by the member number n and divided by two. If in the formula the value of the n-th member is replaced by the expression from the previous paragraph of the article, we get:
Calculation example
For example, let's solve a problem with the following conditions:
The first term of the sequence is zero;
The difference is 0.5.
In the problem, it is required to determine the sum of the terms of the series from 56 to 101.
Solution. Let's use the formula for determining the sum of the progression:
s(n) = (2∙a1 + d∙(n-1))∙n/2
First, we determine the sum of the values of 101 members of the progression by substituting the given conditions of our problem into the formula:
s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2 525
Obviously, in order to find out the sum of the terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.
s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5
So the sum of the arithmetic progression for this example is:
s 101 - s 55 \u003d 2,525 - 742.5 \u003d 1,782.5
Example of practical application of arithmetic progression
At the end of the article, let's return to the example of the arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider such an example.
Getting into a taxi (which includes 3 km) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles / km. Travel distance 30 km. Calculate the cost of the trip.
1. Let's discard the first 3 km, the price of which is included in the landing cost.
30 - 3 = 27 km.
2. Further calculation is nothing more than parsing an arithmetic number series.
The member number is the number of kilometers traveled (minus the first three).
The value of the member is the sum.
The first term in this problem will be equal to a 1 = 50 rubles.
Progression difference d = 22 p.
the number of interest to us - the value of the (27 + 1)th member of the arithmetic progression - the meter reading at the end of the 27th kilometer - 27.999 ... = 28 km.
a 28 \u003d 50 + 22 ∙ (28 - 1) \u003d 644
Calculations of calendar data for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the luminary. In addition, various numerical series are successfully used in statistics and other applied branches of mathematics.
Another kind of number sequence is geometric
A geometric progression is characterized by a large, compared with an arithmetic, rate of change. It is no coincidence that in politics, sociology, medicine, often, in order to show the high speed of the spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops exponentially.
The N-th member of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first member is 1, the denominator is 2, respectively, then:
n=1: 1 ∙ 2 = 2
n=2: 2 ∙ 2 = 4
n=3: 4 ∙ 2 = 8
n=4: 8 ∙ 2 = 16
n=5: 16 ∙ 2 = 32,
b n - the value of the current member of the geometric progression;
b n+1 - the formula of the next member of the geometric progression;
q is the denominator of a geometric progression (constant number).
If the graph of an arithmetic progression is a straight line, then the geometric one draws a slightly different picture:
As in the case of arithmetic, a geometric progression has a formula for the value of an arbitrary member. Any n-th term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:
Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Find the 5th term of the progression
b 5 \u003d b 1 ∙ q (5-1) \u003d 3 ∙ 1.5 4 \u003d 15.1875
The sum of a given number of members is also calculated using a special formula. The sum of the first n members of a geometric progression is equal to the difference between the product of the nth member of the progression and its denominator and the first member of the progression, divided by the denominator reduced by one:
If b n is replaced using the formula discussed above, the value of the sum of the first n members of the considered number series will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set equal to 3. Let's find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280
