The number d is called an arithmetic progression. Arithmetic and geometric progressions

Some people treat the word “progression” with caution, as a very complex term from the branches of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi meter (where they still exist). And understanding the essence (and in mathematics there is nothing more important than “understanding the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.

Mathematical number sequence

A numerical sequence is usually called a series of numbers, each of which has its own number.

a 1 is the first member of the sequence;

and 2 is the second term 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 numbers and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth term is related to its ordinal number by a relationship that can be clearly formulated mathematically. In other words: the numerical value of the nth number is some function of n.

a is the value of a member of a numerical sequence;

n is its serial number;

f(n) is a function, where the ordinal number in the numerical 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 term of an arithmetic sequence is as follows:

a n - the value of the current member of the arithmetic progression;

a n+1 - formula of the next number;

d - difference (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.

Specified member value

Sometimes it is necessary to determine the value of any arbitrary term a n of an arithmetic progression. This can be done by sequentially calculating the values ​​of all members of the arithmetic progression, starting from the first to the desired one. However, this path is not always acceptable if, for example, it is necessary to find the value of the five-thousandth or eight-millionth term. Traditional calculations will take a lot of time. However, a specific arithmetic progression can be studied using certain formulas. There is also a formula for the nth term: the value of any term of an arithmetic progression can be determined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the desired term, reduced by one.

The formula is universal for increasing and decreasing progression.

An example of calculating the value of a given term

Let us solve the following problem of finding the value of the nth term of an arithmetic progression.

Condition: there is an arithmetic progression with parameters:

The first term of the sequence is 3;

The difference in the number series is 1.2.

Task: you need to find the value of 214 terms

Solution: to determine the value of a given term, 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 term of the sequence is equal to 258.6.

The advantages of this method of calculation are obvious - the entire solution takes no more than 2 lines.

Sum of a given number of terms

Very often, in a given arithmetic series, it is necessary to determine the sum of the values ​​of some of its segments. To do this, there is also no need to calculate the values ​​of each term and then add them up. This method is applicable if the number of terms whose sum needs to be found is small. In other cases, it is more convenient to use the following formula.

The sum of the terms of an arithmetic progression from 1 to n is equal to the sum of the first and nth terms, multiplied by the number of the term n and divided by two. If in the formula the value of the nth term 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.

The problem requires determining the sum of the terms of the series from 56 to 101.

Solution. Let's use the formula for determining the amount of progression:

s(n) = (2∙a1 + d∙(n-1))∙n/2

First, we determine the sum of the values ​​of 101 terms 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

Thus, the sum of the arithmetic progression for this example is:

s 101 - s 55 = 2,525 - 742.5 = 1,782.5

Example of practical application of arithmetic progression

At the end of the article, let's return to the example of an arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider this example.

Boarding a taxi (which includes 3 km of travel) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles/km. Travel distance is 30 km. Calculate the cost of the trip.

1. Let’s discard the first 3 km, the price of which is included in the cost of landing.

30 - 3 = 27 km.

2. Further calculation is nothing more than parsing an arithmetic number series.

Member number - 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 r.

the number we are interested in is the value of the (27+1)th term of the arithmetic progression - the meter reading at the end of the 27th kilometer is 27.999... = 28 km.

a 28 = 50 + 22 ∙ (28 - 1) = 644

Calendar data calculations 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 star. In addition, various number series are successfully used in statistics and other applied areas of mathematics.

Another type of number sequence is geometric

Geometric progression is characterized by greater rates of change compared to arithmetic progression. It is no coincidence that in politics, sociology, and medicine, in order to show the high speed of spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops in geometric progression.

The Nth term 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 term is 1, the denominator is correspondingly equal to 2, 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 term of the geometric progression;

b n+1 - formula of the next term of the geometric progression;

q is the denominator of the geometric progression (a constant number).

If the graph of an arithmetic progression is a straight line, then a geometric progression paints a slightly different picture:

As in the case of arithmetic, geometric progression has a formula for the value of an arbitrary term. Any nth 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. Let's find the 5th term of the progression

b 5 = b 1 ∙ q (5-1) = 3 ∙ 1.5 4 = 15.1875

The sum of a given number of terms is also calculated using a special formula. The sum of the first n terms of a geometric progression is equal to the difference between the product of the nth term of the progression and its denominator and the first term 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 terms of the number series under consideration will take the form:

Example. The geometric progression starts with the first term equal to 1. The denominator is set to 3. Let's find the sum of the first eight terms.

s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280

The concept of a number sequence implies that each natural number corresponds to some real value. Such a series of numbers can be either arbitrary or have certain properties - a progression. In the latter case, each subsequent element (member) of the sequence can be calculated using the previous one.

An arithmetic progression is a sequence of numerical values ​​in which its neighboring members differ from each other by the same number (all elements of the series, starting from the 2nd, have a similar property). This number - the difference between the previous and subsequent terms - is constant and is called the progression difference.

Progression difference: definition

Consider a sequence consisting of j values ​​A = a(1), a(2), a(3), a(4) ... a(j), j belongs to the set of natural numbers N. An arithmetic progression, according to its definition, is a sequence , in which a(3) – a(2) = a(4) – a(3) = a(5) – a(4) = … = a(j) – a(j-1) = d. The value d is the desired difference of this progression.

d = a(j) – a(j-1).

Highlight:

• An increasing progression, in which case d > 0. Example: 4, 8, 12, 16, 20, ...
• Decreasing progression, then d< 0. Пример: 18, 13, 8, 3, -2, …

Difference progression and its arbitrary elements

If 2 arbitrary terms of the progression are known (i-th, k-th), then the difference for a given sequence can be determined based on the relationship:

a(i) = a(k) + (i – k)*d, which means d = (a(i) – a(k))/(i-k).

Difference of progression and its first term

This expression will help determine an unknown value only in cases where the number of the sequence element is known.

Progression difference and its sum

The sum of a progression is the sum of its terms. To calculate the total value of its first j elements, use the appropriate formula:

S(j) =((a(1) + a(j))/2)*j, but since a(j) = a(1) + d(j – 1), then S(j) = ((a(1) + a(1) + d(j – 1))/2)*j=(( 2a(1) + d(– 1))/2)*j.

If for every natural number n match a real number a n , then they say that it is given number sequence :

a 1 , a 2 , a 3 , . . . , a n , . . . .

So, the number sequence is a function of the natural argument.

Number a 1 called first term of the sequence , number a 2 — second term of the sequence , number a 3 — third and so on. Number a n called nth member of the sequence , and a natural number n — his number .

From two adjacent members a n And a n +1 sequence member a n +1 called subsequent (towards a n ), A a n — previous (towards a n +1 ).

To define a sequence, you need to specify a method that allows you to find a member of the sequence with any number.

Often the sequence is specified using nth term formulas , that is, a formula that allows you to determine a member of a sequence by its number.

For example,

a 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 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven terms of the numerical sequence are established 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.

Sequence of prime numbers:

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 decreasing , if each of its members, starting from the second, is less than the previous one.

For example,

2, 4, 6, 8, . . . , 2n, . . . — increasing sequence;

1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . — decreasing sequence. ◄

A sequence whose elements do not decrease as the number increases, or, conversely, do not increase, is called monotonous sequence .

Monotonic sequences, in particular, are increasing sequences and decreasing sequences.

Arithmetic progression

Arithmetic progression is a sequence in which each member, 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 the condition is met:

a n +1 = a n + d,

Where d - a certain number.

Thus, the difference between the subsequent and previous terms 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 difference of arithmetic progression.

To define an arithmetic progression, it is enough to indicate its first term and difference.

For example,

If a 1 = 3, d = 4 , then we find the first five terms of the sequence 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 the difference d her n

a n = a 1 + (n- 1)d.

For example,

find the thirtieth term of the 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

Each member of an arithmetic progression, starting from the second, is equal to the arithmetic mean of the preceding and subsequent members.

the numbers a, b and c are successive terms 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 above statement. We have:

a n = 2n- 7,

a n-1 = 2(n- 1) - 7 = 2n- 9,

a n+1 = 2(n+ 1) - 7 = 2n- 5.

Hence,

◄

Note that n The th term 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 down

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

any member of an arithmetic progression, starting from the second, is equal to half the sum of the equally spaced members of this arithmetic progression.

In addition, for any arithmetic progression the following equality holds:

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 terms of an arithmetic progression is equal to the product of half the sum of the extreme terms and the number of terms:

From here, in particular, it follows that if you need 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 connected by two formulas:

Therefore, if the values ​​of 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 is a sequence in which each member, 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 the condition is met:

b n +1 = b n · q,

Where q ≠ 0 - a certain number.

Thus, the ratio of the subsequent term of a given 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 geometric progression.

To define a geometric progression, it is enough to indicate its first term and denominator.

For example,

If b 1 = 1, q = -3 , then we find the first five terms of the sequence 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 The th term can be found using the formula:

b n = b 1 · qn -1 .

For example,

find the seventh term of the geometric progression 1, 2, 4, . . .

b 1 = 1, q = 2,

b 7 = b 1 · q 6 = 1 2 6 = 64. ◄

b n-1 = b 1 · qn -2 ,

b n = b 1 · qn -1 ,

b n +1 = b 1 · qn,

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 preceding and subsequent members.

Since the converse is also true, the following statement holds:

the numbers a, b and c are successive terms of some geometric progression if and only if the square of one of them is equal to the product of 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 above statement. We have:

b n= -3 2 n,

b n -1 = -3 2 n -1 ,

b n +1 = -3 2 n +1 .

Hence,

b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) · (-3 · 2 n +1 ) = b n -1 · b n +1 ,

which proves the desired statement. ◄

Note that n The th term of a geometric progression can be found not only through b 1 , but also any previous member b k , for which it is enough to use the formula

b n = b k · qn - k.

For example,

For b 5 can be written down

b 5 = b 1 · q 4 ,

b 5 = b 2 · q 3,

b 5 = b 3 · q 2,

b 5 = b 4 · q. ◄

b n = b k · qn - k,

b n = b n - k · q k,

then obviously

b n 2 = b n - k· b n + k

the square of any term of a geometric progression, starting from the second, is equal to the product of the terms 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,

in geometric progression

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 denominator q ≠ 0 calculated by the formula:

And when q = 1 - according to the formula

S n= nb 1

Note that if you need to sum the terms

b k, b k +1 , . . . , b n,

then the formula is used:

For example,

in geometric progression 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 a geometric progression is given, then the quantities b 1 , b n, q, n And S n connected 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 properties of monotonicity :

• progression is increasing if one of the following conditions is met:

b 1 > 0 And q> 1;

b 1 < 0 And 0 < q< 1;

• The 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 q< 0 , then the geometric progression is alternating: its terms with odd numbers have the same sign as its first term, and terms with even numbers 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 using the formula:

Pn= 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 called an infinite geometric progression whose denominator modulus is less 1 , that is

|q| < 1 .

Note that an infinitely decreasing geometric progression may not be a decreasing sequence. It fits the occasion

1 < q< 0 .

With such a denominator, the sequence is 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 ones approaches without limit n members of a progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula

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 look at just two examples.

a 1 , a 2 , a 3 , . . . d , That

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 , . . . - geometric progression with denominator 7 2 . ◄

b 1 , b 2 , b 3 , . . . - geometric progression with denominator q , That

log a b 1, log a b 2, log a b 3, . . . - arithmetic progression with difference log aq .

For example,

2, 12, 72, . . . - geometric progression with denominator 6 And

lg 2, lg 12, lg 72, . . . - arithmetic progression with difference lg 6 . ◄

Problems on arithmetic progression existed already in ancient times. They appeared and demanded a solution because they had a practical need.

Thus, one of the papyri of Ancient Egypt that has mathematical content, the Rhind papyrus (19th century BC), contains the following task: divide ten measures of bread among ten people, provided that the difference between each of them is one eighth of the measure.”

And in the mathematical works of the ancient Greeks there are elegant theorems related to arithmetic progression. Thus, Hypsicles of Alexandria (2nd century, who compiled many interesting problems and added the fourteenth book to Euclid’s Elements), formulated the idea: “In an arithmetic progression that has an even number of terms, the sum of the terms of the 2nd half is greater than the sum of the terms of the 1st on the square 1/ 2 numbers of members."

The sequence is denoted by an. The numbers of a sequence are called its members and are usually designated by letters with indices that indicate the serial number of this member (a1, a2, a3 ... read: “a 1st”, “a 2nd”, “a 3rd” and so on ).

The sequence can be infinite or finite.

What is an arithmetic progression? By it we mean the one obtained by adding the previous term (n) with the same number d, which is the difference of the progression.

If d<0, то мы имеем убывающую прогрессию. Если d>0, then this progression is considered increasing.

An arithmetic progression is called finite if only its first few terms are taken into account. With a very large number of members, this is already an endless progression.

Any arithmetic progression is defined by the following formula:

an =kn+b, while b and k are some numbers.

The opposite statement is absolutely true: if a sequence is given by a similar formula, then it is exactly an arithmetic progression that has the properties:

1. Each term of the progression is the arithmetic mean of the previous term and the subsequent one.
2. Converse: if, starting from the 2nd, each term is the arithmetic mean of the previous term and the subsequent one, i.e. if the condition is met, then this sequence is an arithmetic progression. This equality is also a sign of progression, which is why it is usually called a characteristic property of progression.
   In the same way, the theorem that reflects this property is true: a sequence is an arithmetic progression only if this equality is true for any of the terms of the sequence, starting with the 2nd.

The characteristic property for any four numbers of an arithmetic progression can be expressed by the formula an + am = ak + al, if n + m = k + l (m, n, k are progression numbers).

In an arithmetic progression, any necessary (Nth) term can be found using the following formula:

For example: the first term (a1) in an arithmetic progression is given and equal to three, and the difference (d) is equal to four. You need to find the forty-fifth term of this progression. a45 = 1+4(45-1)=177

The formula an = ak + d(n - k) allows you to determine the nth term of an arithmetic progression through any of its kth terms, provided that it is known.

The sum of the terms of an arithmetic progression (meaning the first n terms of a finite progression) is calculated as follows:

Sn = (a1+an) n/2.

If the 1st term is also known, then another formula is convenient for calculation:

Sn = ((2a1+d(n-1))/2)*n.

The sum of an arithmetic progression that contains n terms is calculated as follows:

The choice of formulas for calculations depends on the conditions of the problems and the initial data.

The natural series of any numbers, such as 1,2,3,...,n,..., is the simplest example of an arithmetic progression.

In addition to the arithmetic progression, there is also a geometric progression, which has its own properties and characteristics.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

An arithmetic progression is a series of numbers in which each number is greater (or less) than the previous one by the same amount.

This topic often seems complex and incomprehensible. The indices of the letters, the nth term of the progression, the difference of the progression - all this is somehow confusing, yes... Let's figure out the meaning of the arithmetic progression and everything will get better right away.)

The concept of arithmetic progression.

Arithmetic progression is a very simple and clear concept. Do you have any doubts? In vain.) See for yourself.

I'll write an unfinished series of numbers:

1, 2, 3, 4, 5, ...

Can you extend this series? What numbers will come next, after the five? Everyone... uh..., in short, everyone will realize that the numbers 6, 7, 8, 9, etc. will come next.

Let's complicate the task. I give you an unfinished series of numbers:

2, 5, 8, 11, 14, ...

You will be able to catch the pattern, extend the series, and name seventh row number?

If you realized that this number is 20, congratulations! Not only did you feel key points of arithmetic progression, but also successfully used them in business! If you haven’t figured it out, read on.

Now let’s translate the key points from sensations into mathematics.)

First key point.

Arithmetic progression deals with series of numbers. This is confusing at first. We are used to solving equations, drawing graphs and all that... But here we extend the series, find the number of the series...

It's OK. It’s just that progressions are the first acquaintance with a new branch of mathematics. The section is called "Series" and works specifically with series of numbers and expressions. Get used to it.)

Second key point.

In an arithmetic progression, any number is different from the previous one by the same amount.

In the first example, this difference is one. Whatever number you take, it is one more than the previous one. In the second - three. Any number is three more than the previous one. Actually, it is this moment that gives us the opportunity to grasp the pattern and calculate subsequent numbers.

Third key point.

This moment is not striking, yes... But it is very, very important. Here he is: Each progression number is in its place. There is the first number, there is the seventh, there is the forty-fifth, etc. If you mix them up at random, the pattern will disappear. Arithmetic progression will also disappear. What's left is just a series of numbers.

That's the whole point.

Of course, new terms and designations appear in a new topic. You need to know them. Otherwise you won’t understand the task. For example, you will have to decide something like:

Write down the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.

Inspiring?) Letters, some indexes... And the task, by the way, couldn’t be simpler. You just need to understand the meaning of the terms and designations. Now we will master this matter and return to the task.

Terms and designations.

Arithmetic progression is a series of numbers in which each number is different from the previous one by the same amount.

This quantity is called . Let's look at this concept in more detail.

Arithmetic progression difference.

Arithmetic progression difference is the amount by which any progression number more previous one.

One important point. Please pay attention to the word "more". Mathematically, this means that each progression number is by adding difference of arithmetic progression to the previous number.

To calculate, let's say second numbers of the series, you need to first number add this very difference of an arithmetic progression. For calculation fifth- the difference is necessary add To fourth, well, etc.

Arithmetic progression difference May be positive, then each number in the series will turn out to be real more than the previous one. This progression is called increasing. For example:

8; 13; 18; 23; 28; .....

Here each number is obtained by adding positive number, +5 to the previous one.

The difference may be negative, then each number in the series will be less than the previous one. This progression is called (you won’t believe it!) decreasing.

For example:

8; 3; -2; -7; -12; .....

Here each number is also obtained by adding to the previous one, but already a negative number, -5.

By the way, when working with progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. This helps a lot to navigate the decision, spot your mistakes and correct them before it’s too late.

Arithmetic progression difference usually denoted by the letter d.

How to find d? Very simple. It is necessary to subtract from any number in the series previous number. Subtract. By the way, the result of subtraction is called "difference".)

Let us define, for example, d for increasing arithmetic progression:

2, 5, 8, 11, 14, ...

We take any number in the series that we want, for example, 11. We subtract from it previous number those. 8:

This is the correct answer. For this arithmetic progression, the difference is three.

You can take it any progression number, because for a specific progression d-always the same. At least somewhere at the beginning of the row, at least in the middle, at least anywhere. You cannot take only the very first number. Simply because the very first number no previous one.)

By the way, knowing that d=3, finding the seventh number of this progression is very simple. Let's add 3 to the fifth number - we get the sixth, it will be 17. Let's add three to the sixth number, we get the seventh number - twenty.

Let's define d for descending arithmetic progression:

8; 3; -2; -7; -12; .....

I remind you that, regardless of the signs, to determine d need from any number take away the previous one. Choose any progression number, for example -7. His previous number is -2. Then:

d = -7 - (-2) = -7 + 2 = -5

The difference of an arithmetic progression can be any number: integer, fractional, irrational, any number.

Other terms and designations.

Each number in the series is called member of an arithmetic progression.

Each member of the progression has its own number. The numbers are strictly in order, without any tricks. First, second, third, fourth, etc. For example, in the progression 2, 5, 8, 11, 14, ... two is the first term, five is the second, eleven is the fourth, well, you understand...) Please clearly understand - the numbers themselves can be absolutely anything, whole, fractional, negative, whatever, but numbering of numbers- strictly in order!

How to write a progression in general form? No problem! Each number in a series is written as a letter. To denote an arithmetic progression, the letter is usually used a. The member number is indicated by an index at the bottom right. We write terms separated by commas (or semicolons), like this:

a 1, a 2, a 3, a 4, a 5, .....

a 1- this is the first number, a 3- third, etc. Nothing fancy. This series can be written briefly like this: (a n).

Progressions happen finite and infinite.

Ultimate the progression has a limited number of members. Five, thirty-eight, whatever. But it's a finite number.

Infinite progression - has an infinite number of members, as you might guess.)

You can write the final progression through a series like this, all terms and a dot at the end:

a 1, a 2, a 3, a 4, a 5.

Or like this, if there are many members:

a 1, a 2, ... a 14, a 15.

In the short entry you will have to additionally indicate the number of members. For example (for twenty members), like this:

(a n), n = 20

An infinite progression can be recognized by the ellipsis at the end of the row, as in the examples in this lesson.

Now you can solve the tasks. The tasks are simple, purely for understanding the meaning of an arithmetic progression.

Examples of tasks on arithmetic progression.

Let's look at the task given above in detail:

1. Write out the first six terms of the arithmetic progression (a n), if a 2 = 5, d = -2.5.

We translate the task into understandable language. An infinite arithmetic progression is given. The second number of this progression is known: a 2 = 5. The progression difference is known: d = -2.5. We need to find the first, third, fourth, fifth and sixth terms of this progression.

For clarity, I will write down a series according to the conditions of the problem. The first six terms, where the second term is five:

a 1, 5, a 3, a 4, a 5, a 6,....

a 3 = a 2 + d

Substitute into expression a 2 = 5 And d = -2.5. Don't forget about the minus!

a 3=5+(-2,5)=5 - 2,5 = 2,5

The third term turned out to be smaller than the second. Everything is logical. If the number is greater than the previous one negative value, which means the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We count the fourth term of our series:

a 4 = a 3 + d

a 4=2,5+(-2,5)=2,5 - 2,5 = 0

a 5 = a 4 + d

a 5=0+(-2,5)= - 2,5

a 6 = a 5 + d

a 6=-2,5+(-2,5)=-2,5 - 2,5 = -5

So, terms from the third to the sixth were calculated. The result is the following series:

a 1, 5, 2.5, 0, -2.5, -5, ....

It remains to find the first term a 1 according to the well-known second. This is a step in the other direction, to the left.) So, the difference of the arithmetic progression d should not be added to a 2, A take away:

a 1 = a 2 - d

a 1=5-(-2,5)=5 + 2,5=7,5

That's it. Assignment answer:

7,5, 5, 2,5, 0, -2,5, -5, ...

In passing, I would like to note that we solved this task recurrent way. This terrible word means only the search for a member of the progression according to the previous (adjacent) number. We'll look at other ways to work with progression below.

One important conclusion can be drawn from this simple task.

Remember:

If we know at least one term and the difference of an arithmetic progression, we can find any term of this progression.

Do you remember? This simple conclusion allows you to solve most of the problems of the school course on this topic. All tasks revolve around three main parameters: member of an arithmetic progression, difference of a progression, number of a member of the progression. All.

Of course, all previous algebra is not canceled.) Inequalities, equations, and other things are attached to progression. But according to the progression itself- everything revolves around three parameters.

As an example, let's look at some popular tasks on this topic.

2. Write the finite arithmetic progression as a series if n=5, d = 0.4, and a 1 = 3.6.

Everything is simple here. Everything has already been given. You need to remember how the members of an arithmetic progression are counted, count them, and write them down. It is advisable not to miss the words in the task conditions: “final” and “ n=5". So as not to count until you are completely blue in the face.) There are only 5 (five) members in this progression:

a 2 = a 1 + d = 3.6 + 0.4 = 4

a 3 = a 2 + d = 4 + 0.4 = 4.4

a 4 = a 3 + d = 4.4 + 0.4 = 4.8

a 5 = a 4 + d = 4.8 + 0.4 = 5.2

It remains to write down the answer:

3,6; 4; 4,4; 4,8; 5,2.

Another task:

3. Determine whether the number 7 will be a member of the arithmetic progression (a n), if a 1 = 4.1; d = 1.2.

Hmm... Who knows? How to determine something?

How-how... Write down the progression in the form of a series and see whether there will be a seven there or not! We count:

a 2 = a 1 + d = 4.1 + 1.2 = 5.3

a 3 = a 2 + d = 5.3 + 1.2 = 6.5

a 4 = a 3 + d = 6.5 + 1.2 = 7.7

4,1; 5,3; 6,5; 7,7; ...

Now it is clearly visible that we are just seven slipped through between 6.5 and 7.7! Seven did not fall into our series of numbers, and, therefore, seven will not be a member of the given progression.

Answer: no.

And here is a problem based on a real version of the GIA:

4. Several consecutive terms of the arithmetic progression are written out:

...; 15; X; 9; 6; ...

Here is a series written without end and beginning. No member numbers, no difference d. It's OK. To solve the problem, it is enough to understand the meaning of an arithmetic progression. Let's look and see what's possible to know from this series? What are the three main parameters?

Member numbers? There is not a single number here.

But there are three numbers and - attention! - word "consistent" in condition. This means that the numbers are strictly in order, without gaps. Are there two in this row? neighboring known numbers? Yes, I have! These are 9 and 6. Therefore, we can calculate the difference of the arithmetic progression! Subtract from six previous number, i.e. nine:

There are mere trifles left. What number will be the previous one for X? Fifteen. This means that X can be easily found by simple addition. Add the difference of the arithmetic progression to 15:

That's all. Answer: x=12

We solve the following problems ourselves. Note: these problems are not based on formulas. Purely to understand the meaning of an arithmetic progression.) We just write down a series of numbers and letters, look and figure it out.

5. Find the first positive term of the arithmetic progression if a 5 = -3; d = 1.1.

6. It is known that the number 5.5 is a member of the arithmetic progression (a n), where a 1 = 1.6; d = 1.3. Determine the number n of this member.

7. It is known that in arithmetic progression a 2 = 4; a 5 = 15.1. Find a 3 .

8. Several consecutive terms of the arithmetic progression are written out:

...; 15.6; X; 3.4; ...

Find the term of the progression indicated by the letter x.

9. The train began moving from the station, uniformly increasing speed by 30 meters per minute. What will be the speed of the train after five minutes? Give your answer in km/hour.

10. It is known that in arithmetic progression a 2 = 5; a 6 = -5. Find a 1.

Answers (in disarray): 7.7; 7.5; 9.5; 9; 0.3; 4.

Everything worked out? Amazing! You can master arithmetic progression at a higher level in the following lessons.

Didn't everything work out? No problem. In Special Section 555, all these problems are sorted out piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution to such tasks clearly, clearly, at a glance!

By the way, in the train puzzle there are two problems that people often stumble over. One is purely in terms of progression, and the second is general for any problems in mathematics, and physics too. This is a translation of dimensions from one to another. It shows how these problems should be solved.

In this lesson we looked at the elementary meaning of an arithmetic progression and its main parameters. This is enough to solve almost all problems on this topic. Add d to the numbers, write a series, everything will be solved.

The finger solution works well for very short pieces of a row, as in the examples in this lesson. If the series is longer, the calculations become more complicated. For example, if in problem 9 in the question we replace "five minutes" on "thirty-five minutes" the problem will become significantly worse.)

And there are also tasks that are simple in essence, but absurd in terms of calculations, for example:

An arithmetic progression (a n) is given. Find a 121 if a 1 =3 and d=1/6.

So what, are we going to add 1/6 many, many times?! You can kill yourself!?

You can.) If you don’t know a simple formula by which you can solve such tasks in a minute. This formula will be in the next lesson. And this problem is solved there. In a minute.)

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