Before we start deciding arithmetic progression problems, let's look at what it is number sequence, since an arithmetic progression is a special case of a number sequence.

A number sequence is a number set, each element of which has its own serial number. The elements of this set are called members of the sequence. The serial number of a sequence element is indicated by an index:

The first element of the sequence;

The fifth element of the sequence;

- the “nth” element of the sequence, i.e. element "standing in queue" at number n.

There is a relationship between the value of a sequence element and its sequence number. Therefore, we can consider a sequence as a function whose argument is the ordinal number of the element of the sequence. In other words, we can say that the sequence is a function of the natural argument:

The sequence can be set in three ways:

1 . The sequence can be specified using a table. In this case, we simply set the value of each member of the sequence.

For example, Someone decided to take up personal time management, and to begin with, count how much time he spends on VKontakte during the week. By recording the time in the table, he will receive a sequence consisting of seven elements:

The first line of the table indicates the number of the day of the week, the second - the time in minutes. We see that, that is, on Monday Someone spent 125 minutes on VKontakte, that is, on Thursday - 248 minutes, and, that is, on Friday only 15.

2 . The sequence can be specified using the nth term formula.

In this case, the dependence of the value of a sequence element on its number is expressed directly in the form of a formula.

For example, if , then

To find the value of a sequence element with a given number, we substitute the element number into the formula of the nth term.

We do the same thing if we need to find the value of a function if the value of the argument is known. We substitute the value of the argument into the function equation:

If, for example, , That

Let me note once again that in a sequence, unlike an arbitrary numerical function, the argument can only be a natural number.

3 . The sequence can be specified using a formula that expresses the dependence of the value of the sequence member number n on the values ​​of the previous members. In this case, it is not enough for us to know only the number of the sequence member to find its value. We need to specify the first member or first few members of the sequence.

For example, consider the sequence ,

We can find the values ​​of sequence members one by one, starting from the third:

That is, every time, to find the value of the nth term of the sequence, we return to the previous two. This method of specifying a sequence is called recurrent, from the Latin word recurro- come back.

Now we can define an arithmetic progression. Arithmetic progression is a simple special case of a number sequence.

Arithmetic progression is a numerical sequence, each member of which, starting from the second, is equal to the previous one added to the same number.

The number is called difference of arithmetic progression. The difference of an arithmetic progression can be positive, negative, or equal to zero.

If title="d>0">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} increasing.

For example, 2; 5; 8; 11;...

If , then each term of an arithmetic progression is less than the previous one, and the progression is decreasing.

For example, 2; -1; -4; -7;...

If , then all terms of the progression are equal to the same number, and the progression is stationary.

For example, 2;2;2;2;...

The main property of an arithmetic progression:

Let's look at the picture.

We see that

, and at the same time

Adding these two equalities, we get:

.

Divide both sides of the equality by 2:

So, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the two neighboring ones:

Moreover, since

, and at the same time

, That

, and therefore

Each term of an arithmetic progression, starting with title="k>l">, равен среднему арифметическому двух равноотстоящих. !}

Formula of the th term.

We see that the terms of the arithmetic progression satisfy the following relations:

and finally

We got formula of the nth term.

IMPORTANT! Any member of an arithmetic progression can be expressed through and. Knowing the first term and the difference of an arithmetic progression, you can find any of its terms.

The sum of n terms of an arithmetic progression.

In an arbitrary arithmetic progression, the sums of terms equidistant from the extreme ones are equal to each other:

Consider an arithmetic progression with n terms. Let the sum of n terms of this progression be equal to .

Let's arrange the terms of the progression first in ascending order of numbers, and then in descending order:

Let's add in pairs:

The sum in each bracket is , the number of pairs is n.

We get:

So, the sum of n terms of an arithmetic progression can be found using the formulas:

Let's consider solving arithmetic progression problems.

1 . The sequence is given by the formula of the nth term: . Prove that this sequence is an arithmetic progression.

Let us prove that the difference between two adjacent terms of the sequence is equal to the same number.

We found that the difference between two adjacent members of the sequence does not depend on their number and is a constant. Therefore, by definition, this sequence is an arithmetic progression.

2 . Given an arithmetic progression -31; -27;...

a) Find 31 terms of the progression.

b) Determine whether the number 41 is included in this progression.

A) We see that ;

Let's write down the formula for the nth term for our progression.

In general

In our case , That's why

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

So, in one of the papyri Ancient Egypt", which has a 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. At very large quantities members 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 us to determine nth term 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 1st 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.

Natural series of any numbers, such as 1,2,3,...,n,...- simplest example arithmetic progression.

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

Sum of an arithmetic progression.

The sum of an arithmetic progression is a simple thing. Both in meaning and in formula. But there are all sorts of tasks on this topic. From basic to quite solid.

First, let's understand the meaning and formula of the amount. And then we'll decide. For your own pleasure.) The meaning of the amount is as simple as a moo. To find the sum of an arithmetic progression, you just need to carefully add all its terms. If these terms are few, you can add without any formulas. But if there is a lot, or a lot... addition is annoying.) In this case, the formula comes to the rescue.

The formula for the amount is simple:

Let's figure out what kind of letters are included in the formula. This will clear things up a lot.

S n - the sum of an arithmetic progression. Addition result everyone members, with first By last. This is important. They add up exactly All members in a row, without skipping or skipping. And, precisely, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of the fifth to twentieth terms, direct application of the formula will disappoint.)

a 1 - first member of the progression. Everything is clear here, it's simple first row number.

a n- last member of the progression. The last number of the series. Not a very familiar name, but when applied to the amount, it’s very suitable. Then you will see for yourself.

n - number of the last member. It is important to understand that in the formula this number coincides with the number of terms added.

Let's define the concept last member a n. Tricky question: which member will be the last one if given endless arithmetic progression?)

To answer confidently, you need to understand the elementary meaning of an arithmetic progression and... read the task carefully!)

In the task of finding the sum of an arithmetic progression, the last term always appears (directly or indirectly), which should be limited. Otherwise, a final, specific amount simply doesn't exist. For the solution, it does not matter whether the progression is given: finite or infinite. It doesn’t matter how it is given: a series of numbers, or a formula for the nth term.

The most important thing is to understand that the formula works from the first term of the progression to the term with number n. Actually, the full name of the formula looks like this: the sum of the first n terms of an arithmetic progression. The number of these very first members, i.e. n, is determined solely by the task. In a task, all this valuable information is often encrypted, yes... But never mind, in the examples below we reveal these secrets.)

Examples of tasks on the sum of an arithmetic progression.

First of all, useful information:

The main difficulty in tasks involving the sum of an arithmetic progression lies in the correct determination of the elements of the formula.

The task writers encrypt these very elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough to simply decipher them. Let's look at a few examples in detail. Let's start with a task based on a real GIA.

1. The arithmetic progression is given by the condition: a n = 2n-3.5. Find the sum of its first 10 terms.

Good job. Easy.) To determine the amount using the formula, what do we need to know? First member a 1, last term a n, yes the number of the last member n.

Where can I get the last member's number? n? Yes, right there, on condition! It says: find the sum first 10 members. Well, what number will it be with? last, tenth member?) You won’t believe it, his number is tenth!) Therefore, instead of a n we will substitute into the formula a 10, and instead n- ten. I repeat, the number of the last member coincides with the number of members.

It remains to determine a 1 And a 10. This is easily calculated using the formula for the nth term, which is given in the problem statement. Don't know how to do this? Attend the previous lesson, without this there is no way.

a 1= 2 1 - 3.5 = -1.5

a 10=2·10 - 3.5 =16.5

S n = S 10.

We have found out the meaning of all elements of the formula for the sum of an arithmetic progression. All that remains is to substitute them and count:

That's it. Answer: 75.

Another task based on the GIA. A little more complicated:

2. Given an arithmetic progression (a n), the difference of which is 3.7; a 1 =2.3. Find the sum of its first 15 terms.

We immediately write the sum formula:

This formula allows us to find the value of any term by its number. We look for a simple substitution:

a 15 = 2.3 + (15-1) 3.7 = 54.1

It remains to substitute all the elements into the formula for the sum of an arithmetic progression and calculate the answer:

Answer: 423.

By the way, if in the sum formula instead of a n We simply substitute the formula for the nth term and get:

Let us present similar ones and obtain a new formula for the sum of terms of an arithmetic progression:

As you can see, the nth term is not required here a n. In some problems this formula helps a lot, yes... You can remember this formula. Or you can simply display it at the right time, like here. After all, you always need to remember the formula for the sum and the formula for the nth term.)

Now the task in the form of a short encryption):

3. Find the sum of all positive two-digit numbers that are multiples of three.

Wow! Neither your first member, nor your last, nor progression at all... How to live!?

You will have to think with your head and pull out all the elements of the sum of the arithmetic progression from the condition. We know what two-digit numbers are. They consist of two numbers.) What two-digit number will be first? 10, presumably.) A last double digit number? 99, of course! The three-digit ones will follow him...

Multiples of three... Hm... These are numbers that are divisible by three, here! Ten is not divisible by three, 11 is not divisible... 12... is divisible! So, something is emerging. You can already write down a series according to the conditions of the problem:

12, 15, 18, 21, ... 96, 99.

Will this series be an arithmetic progression? Certainly! Each term differs from the previous one by strictly three. If you add 2 or 4 to a term, say, the result, i.e. the new number is no longer divisible by 3. You can immediately determine the difference of the arithmetic progression: d = 3. It will come in handy!)

So, we can safely write down some progression parameters:

What will the number be? n last member? Anyone who thinks that 99 is fatally mistaken... The numbers always go in a row, but our members jump over three. They don't match.

There are two solutions here. One way is for the super hardworking. You can write down the progression, the entire series of numbers, and count the number of members with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If we apply the formula to our problem, we find that 99 is the thirtieth term of the progression. Those. n = 30.

Let's look at the formula for the sum of an arithmetic progression:

We look and rejoice.) We pulled out from the problem statement everything necessary to calculate the amount:

a 1= 12.

a 30= 99.

S n = S 30.

All that remains is elementary arithmetic. We substitute the numbers into the formula and calculate:

Answer: 1665

Another type of popular puzzle:

4. Given an arithmetic progression:

-21,5; -20; -18,5; -17; ...

Find the sum of terms from twentieth to thirty-four.

We look at the formula for the amount and... we get upset.) The formula, let me remind you, calculates the amount from the first member. And in the problem you need to calculate the sum since the twentieth... The formula won't work.

You can, of course, write out the entire progression in a series, and add terms from 20 to 34. But... it’s somehow stupid and takes a long time, right?)

There are more elegant solution. Let's divide our series into two parts. The first part will be from the first term to the nineteenth. Second part - from twenty to thirty-four. It is clear that if we calculate the sum of the terms of the first part S 1-19, let's add it with the sum of the terms of the second part S 20-34, we get the sum of the progression from the first term to the thirty-fourth S 1-34. Like this:

S 1-19 + S 20-34 = S 1-34

From this we can see that find the sum S 20-34 can be done by simple subtraction

S 20-34 = S 1-34 - S 1-19

Both amounts on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Let's get started?

We extract the progression parameters from the problem statement:

d = 1.5.

a 1= -21,5.

To calculate the sums of the first 19 and first 34 terms, we will need the 19th and 34th terms. We calculate them using the formula for the nth term, as in problem 2:

a 19= -21.5 +(19-1) 1.5 = 5.5

a 34= -21.5 +(34-1) 1.5 = 28

There's nothing left. From the sum of 34 terms subtract the sum of 19 terms:

S 20-34 = S 1-34 - S 1-19 = 110.5 - (-152) = 262.5

Answer: 262.5

One important note! There is a very useful trick in solving this problem. Instead of direct calculation what you need (S 20-34), we counted something that would seem not to be needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the complete result. This kind of “feint with your ears” often saves you in wicked problems.)

In this lesson we looked at problems for which it is enough to understand the meaning of the sum of an arithmetic progression. Well, you need to know a couple of formulas.)

Practical advice:

When solving any problem involving the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.

Formula for the nth term:

These formulas will immediately tell you what to look for and in what direction to think in order to solve the problem. Helps.

And now the tasks for independent solution.

5. Find the sum of all two-digit numbers that are not divisible by three.

Cool?) The hint is hidden in the note to problem 4. Well, problem 3 will help.

6. The arithmetic progression is given by the condition: a 1 = -5.5; a n+1 = a n +0.5. Find the sum of its first 24 terms.

Unusual?) This is a recurrent formula. You can read about it in the previous lesson. Don’t ignore the link, such problems are often found in the State Academy of Sciences.

7. Vasya saved up money for the holiday. As much as 4550 rubles! And I decided to give my favorite person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and on each subsequent day spend 50 rubles more than the previous one! Until the money runs out. How many days of happiness did Vasya have?

Difficult?) An additional formula from problem 2 will help.

Answers (in disarray): 7, 3240, 6.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

Arithmetic and geometric progressions

Theoretical information

Theoretical information

	Arithmetic progression	Geometric progression
Definition	Arithmetic progression a n is a sequence in which each member, starting from the second, is equal to the previous member added to the same number d (d- progression difference)	Geometric progression b n is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression)
Recurrence formula	For any natural n a n + 1 = a n + d	For any natural n b n + 1 = b n ∙ q, b n ≠ 0
Formula nth term	a n = a 1 + d (n – 1)	b n = b 1 ∙ q n - 1 , b n ≠ 0
Characteristic property
Sum of the first n terms

Examples of tasks with comments

Task 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+ 21 d

According to the condition:

a 1= -6, then a 22= -6 + 21 d .

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 method (using the n-term formula)

According to the formula for the nth term of a geometric progression:

b 5 = b 1 ∙ q 5 - 1 = b 1 ∙ q 4.

Because b 1 = -3,

2nd method (using recurrent 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 .

From this it follows:

.

Let's substitute the data into 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 is in in this case more convenient to use?

By condition, the formula for the nth term of the original progression is known ( a n) a n= 3n - 4. You can find immediately and a 1, And a 16 without finding d. Therefore, we will 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 the geometric progression are written:

Find the term of the progression labeled x.

When solving, we will use the formula for the nth term b n = b 1 ∙ q n - 1 for geometric progressions. The first term of the progression. To find the denominator of the progression q, you need to take any of the given terms of the progression and divide by the previous one. In our example, we can take and divide by. We obtain that q = 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 the arithmetic progressions given by the formula of the nth term, select the one for which the condition is satisfied a 27 > 9:

Since the given 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 holds a n > -6.

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

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

a n=	a n-1 + a n+1
	2

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,

a n+1 + a n-1	=	2n- 5 + 2n- 9	= 2n- 7 = a n,
2		2

◄

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

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 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 meanings 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. In this case:

• 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:

S n- S k -1 = b k + b k +1 + . . . + b n = b k ·	1 - qn - k +1	.
	1 - q

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

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 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 . ◄