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

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

So, a numerical sequence is a function of a 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 of the sequence , 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,

the 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.

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 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 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- Sk -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 a geometric progression is given, 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 . ◄

First level

Arithmetic progression. Detailed theory with examples (2019)

Numeric sequence

So let's sit down and start writing some numbers. For example:
You can write any numbers, and there can be as many as you like (in our case, them). No matter how many numbers we write, we can always say which of them is the first, which is the second, and so on to the last, that is, we can number them. This is an example of a number sequence:

Numeric sequence
For example, for our sequence:

The assigned number is specific to only one sequence number. In other words, there are no three second numbers in the sequence. The second number (like the -th number) is always the same.
The number with the number is called the -th member of the sequence.

We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .

In our case:

Let's say we have a numerical sequence in which the difference between adjacent numbers is the same and equal.
For example:

etc.
Such a numerical sequence is called an arithmetic progression.
The term "progression" was introduced by the Roman author Boethius as early as the 6th century and was understood in a broader sense as an endless numerical sequence. The name "arithmetic" was transferred from the theory of continuous proportions, which the ancient Greeks were engaged in.

This is a numerical sequence, each member of which is equal to the previous one, added with the same number. This number is called the difference of an arithmetic progression and is denoted.

Try to determine which number sequences are an arithmetic progression and which are not:

a)
b)
c)
d)

Got it? Compare our answers:
Is arithmetic progression - b, c.
Is not arithmetic progression - a, d.

Let's return to the given progression () and try to find the value of its th member. Exists two way to find it.

1. Method

We can add to the previous value of the progression number until we reach the th term of the progression. It’s good that we don’t have much to summarize - only three values:

So, the -th member of the described arithmetic progression is equal to.

2. Method

What if we needed to find the value of the th term of the progression? The summation would have taken us more than one hour, and it is not a fact that we would not have made mistakes when adding the numbers.
Of course, mathematicians have come up with a way in which you do not need to add the difference of an arithmetic progression to the previous value. Look closely at the drawn picture ... Surely you have already noticed a certain pattern, namely:

For example, let's see what makes up the value of the -th member of this arithmetic progression:


In other words:

Try to independently find in this way the value of a member of this arithmetic progression.

Calculated? Compare your entries with the answer:

Pay attention that you got exactly the same number as in the previous method, when we successively added the members of an arithmetic progression to the previous value.
Let's try to "depersonalize" this formula - we bring it into a general form and get:

Arithmetic progression equation.

Arithmetic progressions are either increasing or decreasing.

Increasing- progressions in which each subsequent value of the terms is greater than the previous one.
For example:

Descending- progressions in which each subsequent value of the terms is less than the previous one.
For example:

The derived formula is used in the calculation of terms in both increasing and decreasing terms of an arithmetic progression.
Let's check it out in practice.
We are given an arithmetic progression consisting of the following numbers:

Since then:

Thus, we were convinced that the formula works both in decreasing and in increasing arithmetic progression.
Try to find the -th and -th members of this arithmetic progression on your own.

Let's compare the results:

Arithmetic progression property

Let's complicate the task - we derive the property of an arithmetic progression.
Suppose we are given the following condition:
- arithmetic progression, find the value.
It's easy, you say, and start counting according to the formula you already know:

Let, a, then:

Absolutely right. It turns out that we first find, then add it to the first number and get what we are looking for. If the progression is represented by small values, then there is nothing complicated about it, but what if we are given numbers in the condition? Agree, there is a possibility of making mistakes in the calculations.
Now think, is it possible to solve this problem in one step using any formula? Of course, yes, and we will try to bring it out now.

Let's denote the desired term of the arithmetic progression as, we know the formula for finding it - this is the same formula that we derived at the beginning:
, then:

• the previous member of the progression is:
• the next term of the progression is:

Let's sum the previous and next members of the progression:

It turns out that the sum of the previous and subsequent members of the progression is twice the value of the member of the progression located between them. In other words, in order to find the value of a progression member with known previous and successive values, it is necessary to add them and divide by.

That's right, we got the same number. Let's fix the material. Calculate the value for the progression yourself, because it is not difficult at all.

Well done! You know almost everything about progression! It remains to find out only one formula, which, according to legend, one of the greatest mathematicians of all time, the "king of mathematicians" - Karl Gauss, easily deduced for himself ...

When Carl Gauss was 9 years old, the teacher, busy checking the work of students from other classes, asked the following task at the lesson: "Calculate the sum of all natural numbers from up to (according to other sources up to) inclusive." What was the surprise of the teacher when one of his students (it was Karl Gauss) after a minute gave the correct answer to the task, while most of the classmates of the daredevil after long calculations received the wrong result ...

Young Carl Gauss noticed a pattern that you can easily notice.
Let's say we have an arithmetic progression consisting of -ti members: We need to find the sum of the given members of the arithmetic progression. Of course, we can manually sum all the values, but what if we need to find the sum of its terms in the task, as Gauss was looking for?

Let's depict the progression given to us. Look closely at the highlighted numbers and try to perform various mathematical operations with them.


Tried? What did you notice? Correctly! Their sums are equal

Now answer, how many such pairs will there be in the progression given to us? Of course, exactly half of all numbers, that is.
Based on the fact that the sum of two members of an arithmetic progression is equal, and similar equal pairs, we get that total amount is equal to:
.
Thus, the formula for the sum of the first terms of any arithmetic progression will be:

In some problems, we do not know the th term, but we know the progression difference. Try to substitute in the sum formula, the formula of the th member.
What did you get?

Well done! Now let's return to the problem that was given to Carl Gauss: calculate for yourself what the sum of numbers starting from the -th is, and the sum of the numbers starting from the -th.

How much did you get?
Gauss turned out that the sum of the terms is equal, and the sum of the terms. Is that how you decided?

In fact, the formula for the sum of members of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people used the properties of an arithmetic progression with might and main.
For example, imagine Ancient Egypt and the largest construction site of that time - the construction of a pyramid ... The figure shows one side of it.

Where is the progression here you say? Look carefully and find a pattern in the number of sand blocks in each row of the pyramid wall.


Why not an arithmetic progression? Count how many blocks are needed to build one wall if block bricks are placed in the base. I hope you will not count by moving your finger across the monitor, do you remember the last formula and everything we said about arithmetic progression?

AT this case the progression looks like this:
Arithmetic progression difference.
The number of members of an arithmetic progression.
Let's substitute our data into the last formulas (we count the number of blocks in 2 ways).

Method 1.

Method 2.

And now you can also calculate on the monitor: compare the obtained values ​​​​with the number of blocks that are in our pyramid. Did it agree? Well done, you have mastered the sum of the th terms of an arithmetic progression.
Of course, you can’t build a pyramid from the blocks at the base, but from? Try to calculate how many sand bricks are needed to build a wall with this condition.
Did you manage?
The correct answer is blocks:

Workout

Tasks:

1. Masha is getting in shape for the summer. Every day she increases the number of squats by. How many times will Masha squat in weeks if she did squats at the first workout.
2. What is the sum of all odd numbers contained in.
3. When storing logs, lumberjacks stack them in such a way that each upper layer contains one log less than the previous one. How many logs are in one masonry, if the base of the masonry is logs.

Answers:

1. Let us define the parameters of the arithmetic progression. In this case
   (weeks = days).

   Answer: In two weeks, Masha should squat once a day.

2. First odd number, last number.
   Arithmetic progression difference.
   The number of odd numbers in - half, however, check this fact using the formula for finding the -th member of an arithmetic progression:

   The numbers do contain odd numbers.
   We substitute the available data into the formula:

   Answer: The sum of all odd numbers contained in is equal to.

3. Recall the problem about the pyramids. For our case, a , since each top layer is reduced by one log, there are only a bunch of layers, that is.
   Substitute the data in the formula:

   Answer: There are logs in the masonry.

Summing up

1. - a numerical sequence in which the difference between adjacent numbers is the same and equal. It is increasing and decreasing.
2. Finding formula th member of an arithmetic progression is written by the formula - , where is the number of numbers in the progression.
3. Property of members of an arithmetic progression- - where - the number of numbers in the progression.
4. The sum of the members of an arithmetic progression can be found in two ways:
   
   , where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Numeric sequence

Let's sit down and start writing some numbers. For example:

You can write any numbers, and there can be as many as you like. But you can always tell which of them is the first, which is the second, and so on, that is, we can number them. This is an example of a number sequence.

Numeric sequence is a set of numbers, each of which can be assigned a unique number.

In other words, each number can be associated with a certain natural number, and only one. And we will not assign this number to any other number from this set.

The number with the number is called the -th member of the sequence.

We usually call the whole sequence some letter (for example,), and each member of this sequence - the same letter with an index equal to the number of this member: .

It is very convenient if the -th member of the sequence can be given by some formula. For example, the formula

sets the sequence:

And the formula is the following sequence:

For example, an arithmetic progression is a sequence (the first term here is equal, and the difference). Or (, difference).

nth term formula

We call recurrent a formula in which, in order to find out the -th term, you need to know the previous or several previous ones:

To find, for example, the th term of the progression using such a formula, we have to calculate the previous nine. For example, let. Then:

Well, now it's clear what the formula is?

In each line, we add to, multiplied by some number. For what? Very simple: this is the number of the current member minus:

Much more comfortable now, right? We check:

Decide for yourself:

In an arithmetic progression, find the formula for the nth term and find the hundredth term.

Solution:

The first member is equal. And what is the difference? And here's what:

(after all, it is called the difference because it is equal to the difference of successive members of the progression).

So the formula is:

Then the hundredth term is:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Carl Gauss, being a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last number is equal, the sum of the second and penultimate is the same, the sum of the third and the 3rd from the end is the same, and so on. How many such pairs are there? That's right, exactly half the number of all numbers, that is. So,

The general formula for the sum of the first terms of any arithmetic progression will be:

Example:
Find the sum of all two-digit numbers, multiples.

Solution:

The first such number is this. Each next is obtained by adding a number to the previous one. Thus, the numbers of interest to us form an arithmetic progression with the first term and the difference.

The formula for the th term for this progression is:

How many terms are in the progression if they must all be two digits?

Very easy: .

The last term of the progression will be equal. Then the sum:

Answer: .

Now decide for yourself:

1. Every day the athlete runs 1m more than the previous day. How many kilometers will he run in weeks if he ran km m on the first day?
2. A cyclist rides more miles each day than the previous one. On the first day he traveled km. How many days does he have to drive to cover a kilometer? How many kilometers will he travel on the last day of the journey?
3. The price of a refrigerator in the store is reduced by the same amount every year. Determine how much the price of a refrigerator decreased every year if, put up for sale for rubles, six years later it was sold for rubles.

Answers:

1. The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
   .
   Answer:
2. Here it is given:, it is necessary to find.
   Obviously, you need to use the same sum formula as in the previous problem:
   .
   Substitute the values:

   The root obviously doesn't fit, so the answer.
   Let's calculate the distance traveled over the last day using the formula of the -th member:
   (km).
   Answer:

3. Given: . Find: .
   It doesn't get easier:
   (rub).
   Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN

This is a numerical sequence in which the difference between adjacent numbers is the same and equal.

Arithmetic progression is increasing () and decreasing ().

For example:

The formula for finding the n-th member of an arithmetic progression

is written as a formula, where is the number of numbers in the progression.

Property of members of an arithmetic progression

It makes it easy to find a member of the progression if its neighboring members are known - where is the number of numbers in the progression.

The sum of the members of an arithmetic progression

There are two ways to find the sum:

Where is the number of values.

Where is the number of values.

Attention!
There are additional
material in Special Section 555.
For those who strongly "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 is often difficult and incomprehensible. Letter indexes, nth term progressions, the difference in progression - all this is somehow confusing, yes ... Let's figure out the meaning of the arithmetic progression and everything will work out right away.)

The concept of arithmetic progression.

Arithmetic progression is a very simple and clear concept. Doubt? In vain.) See for yourself.

I'll write an unfinished series of numbers:

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

Can you extend this line? What numbers will go next, after the five? Everyone ... uh ..., in short, everyone will figure out that the numbers 6, 7, 8, 9, etc. will go further.

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

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

You can catch the pattern, extend the series, and name seventh row number?

If you figured out that this number is 20 - I congratulate you! You not only felt key points arithmetic progression, but also successfully used them in business! If you don't understand, 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, building graphs and all that ... And then 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 with series of numbers and expressions. Get used to it.)

Second key point.

In an arithmetic progression, any number differs 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 times greater than the previous one. Actually, it is this moment that gives us the opportunity to catch the pattern and calculate the subsequent numbers.

Third key point.

This moment is not striking, yes ... But very, very important. Here he is: each progression number stands in its place. There is the first number, there is the seventh, there is the forty-fifth, and so on. If you confuse them haphazardly, the pattern will disappear. The arithmetic progression will also disappear. It's just a series of numbers.

That's the whole point.

Of course, in new topic new terms and notation appear. They need to know. Otherwise, you won't understand the task. For example, you have to decide something like:

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

Does it inspire?) Letters, some indexes... And the task, by the way, couldn't be easier. You just need to understand the meaning of the terms and notation. 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 value is called . Let's deal with this concept in more detail.

Arithmetic progression difference.

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

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

To calculate, let's say second numbers of the row, it is necessary 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 of 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 adding positive number, +5 to the previous one.

The difference can 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 every number is obtained too adding to the previous, but already negative number, -5.

By the way, when working with a progression, it is very useful to immediately determine its nature - whether it is increasing or decreasing. It helps a lot to find your bearings in the decision, to detect 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 of the series previous number. Subtract. By the way, the result of subtraction is called "difference".)

Let's define, for example, d for an increasing arithmetic progression:

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

We take any number of the row that we want, for example, 11. Subtract from it the previous number those. eight:

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

You can just take any number of progressions, 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 can not take only the very first number. Just because the very first number no previous.)

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

Let's define d for a decreasing arithmetic progression:

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

I remind you that, regardless of the signs, to determine d needed from any number take away the previous one. We choose any number of progression, 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.

Other terms and designations.

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

Each member of the progression has his 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 member, five is the second, eleven is the fourth, well, you understand ...) Please clearly understand - the numbers themselves can be absolutely any, whole, fractional, negative, whatever, but numbering- strictly in order!

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

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

a 1 is the first number a 3- third, etc. Nothing tricky. You can write this series briefly like this: (a n).

There are progressions finite and infinite.

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

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

burn final progression through a series you can like this, all members 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 a 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 already solve tasks. The tasks are simple, purely for understanding the meaning of the arithmetic progression.

Examples of tasks for arithmetic progression.

Let's take a closer look at the task above:

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

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

For clarity, I will write down a series according to the condition of the problem. The first six members, where the second member is five:

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

a 3 = a 2 + d

We substitute in the expression a 2 = 5 and d=-2.5. Don't forget the minus!

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

The third term is less than the second. Everything is logical. If the number is greater than the previous one negative value, so the number itself will be less than the previous one. Progression is decreasing. Okay, let's take it into account.) We consider the fourth member 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, the terms from the third to the sixth have been calculated. This resulted in a 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.) Hence, 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 all there is to it. Task response:

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

In passing, I note that we solved this task recurrent way. This terrible word means, only, the search for a member of the progression by the previous (adjacent) number. Other ways to work with progression will be discussed later.

One important conclusion can be drawn from this simple task.

Remember:

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

Remember? This simple conclusion allows us 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 a progression. Everything.

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

For example, consider some popular tasks on this topic.

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

Everything is simple here. Everything is already given. You need to remember how the members of an arithmetic progression are calculated, count, and write down. It is advisable not to skip the words in the task condition: "final" and " n=5". In order not to count until you are completely blue in the face.) There are only 5 (five) members in this progression:

a 2 \u003d a 1 + d \u003d 3.6 + 0.4 \u003d 4

a 3 \u003d a 2 + d \u003d 4 + 0.4 \u003d 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 if the number 7 will be a member of an arithmetic progression (a n) if a 1 \u003d 4.1; d = 1.2.

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

How-how ... Yes, write down the progression in the form of a series and see if there will be a seven or not! We believe:

a 2 \u003d a 1 + d \u003d 4.1 + 1.2 \u003d 5.3

a 3 \u003d a 2 + d \u003d 5.3 + 1.2 \u003d 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 seen that we are just seven slipped through between 6.5 and 7.7! The seven did not get into our series of numbers, and, therefore, the seven will not be a member of the given progression.

Answer: no.

And here is a problem based on real version GIA:

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

...; fifteen; X; 9; 6; ...

Here is a series 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 see and see what we can to know from this line? What are the parameters of the three main ones?

Member numbers? There is not a single number here.

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

There are empty spaces left. What number will be the previous one for x? Fifteen. So x can be easily found by simple addition. To 15 add the difference of an arithmetic progression:

That's all. Answer: x=12

We solve the following problems ourselves. Note: these puzzles are not for formulas. Purely for understanding the meaning of an arithmetic progression.) We just write down a series of numbers-letters, look and think.

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 term.

7. It is known that in an arithmetic progression a 2 = 4; a 5 \u003d 15.1. Find a 3 .

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

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

Find the term of the progression, denoted by the letter x.

9. The train started moving from the station, gradually increasing its speed by 30 meters per minute. What will be the speed of the train in five minutes? Give your answer in km/h.

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

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

Everything worked out? Wonderful! You can master the arithmetic progression for more high level, in the next lessons.

Didn't everything work out? No problem. In Special Section 555, all these puzzles are broken down piece by piece.) And, of course, a simple practical technique is described that immediately highlights the solution of such tasks clearly, clearly, as in the palm of your hand!

By the way, in the puzzle about the train there are two problems on which people often stumble. One - purely by progression, and the second - common to any tasks 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 examined 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 decided.

The finger solution works well for very short pieces of the series, 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, replace "five minutes" on the "thirty-five minutes" the problem will become much worse.)

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

Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.

And what, we will add 1/6 many, many times?! Is it possible to kill yourself!?

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

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. Learning - with interest!)

you can get acquainted with functions and derivatives.

Online calculator.
Arithmetic progression solution.
Given: a n , d, n
Find: a 1

This math program finds \(a_1\) of an arithmetic progression based on user-specified numbers \(a_n, d \) and \(n \).
The numbers \(a_n\) and \(d \) can be specified not only as integers, but also as fractions. Moreover, a fractional number can be entered in the form of a decimal fraction (\ (2.5 \)) and in the form common fraction(\(-5\frac(2)(7) \)).

The program not only gives the answer to the problem, but also displays the process of finding a solution.

This online calculator can be useful for high school students general education schools in preparation for control work and exams, when testing knowledge before the exam, parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.

In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.

If you are not familiar with the rules for entering numbers, we recommend that you familiarize yourself with them.

Rules for entering numbers

The numbers \(a_n\) and \(d \) can be specified not only as integers, but also as fractions.
The number \(n\) can only be a positive integer.

Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a dot or a comma.
For example, you can enter decimals so 2.5 or so 2.5

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When you enter numeric fraction The numerator is separated from the denominator by a division sign: /
Input:
Result: \(-\frac(2)(3) \)

The integer part is separated from the fraction by an ampersand: &
Input:
Result: \(-1\frac(2)(3) \)

Enter numbers a n , d, n

Find a 1

It was found that some scripts needed to solve this task were not loaded, and the program may not work.
You may have AdBlock enabled.
In this case, disable it and refresh the page.

You have JavaScript disabled in your browser.
JavaScript must be enabled for the solution to appear.
Here are instructions on how to enable JavaScript in your browser.

Because There are a lot of people who want to solve the problem, your request is queued.
After a few seconds, the solution will appear below.
Wait, please sec...

If you noticed an error in the solution, then you can write about it in the Feedback Form .
Do not forget indicate which task you decide what enter in the fields.

Our games, puzzles, emulators:

A bit of theory.

Numeric sequence

In everyday practice, the numbering of various objects is often used to indicate the order in which they are located. For example, the houses on each street are numbered. In the library, reader's subscriptions are numbered and then arranged in the order of the assigned numbers in special file cabinets.

In a savings bank, by the number of the depositor's personal account, you can easily find this account and see what kind of deposit it has. Let there be a deposit of a1 rubles on account No. 1, a deposit of a2 rubles on account No. 2, etc. It turns out numerical sequence
a 1 , a 2 , a 3 , ..., a N
where N is the number of all accounts. Here, each natural number n from 1 to N is assigned a number a n .

Mathematics also studies infinite number sequences:
a 1 , a 2 , a 3 , ..., a n , ... .
The number a 1 is called the first member of the sequence, number a 2 - the second member of the sequence, number a 3 - the third member of the sequence etc.
The number a n is called nth (nth) member of the sequence, and the natural number n is its number.

For example, in the sequence of squares of natural numbers 1, 4, 9, 16, 25, ..., n 2 , (n + 1) 2 , ... and 1 = 1 is the first member of the sequence; and n = n 2 is the nth member of the sequence; a n+1 = (n + 1) 2 is the (n + 1)th (en plus the first) member of the sequence. Often a sequence can be specified by the formula of its nth member. For example, the formula \(a_n=\frac(1)(n), \; n \in \mathbb(N) \) gives the sequence \(1, \; \frac(1)(2) , \; \frac( 1)(3) , \; \frac(1)(4) , \dots,\frac(1)(n) , \dots \)

Arithmetic progression

The length of a year is approximately 365 days. More exact value equals \(365\frac(1)(4) \) days, so every four years an error of one day accumulates.

To account for this error, a day is added to every fourth year, and the elongated year is called a leap year.

For example, in the third millennium leap years the years are 2004, 2008, 2012, 2016, ... .

In this sequence, each member, starting from the second, is equal to the previous one, added with the same number 4. Such sequences are called arithmetic progressions.

Definition.
The numerical sequence a 1 , a 2 , a 3 , ..., a n , ... is called arithmetic progression, if for all natural n the equality
\(a_(n+1) = a_n+d, \)
where d is some number.

It follows from this formula that a n+1 - a n = d. The number d is called the difference arithmetic progression.

By definition of an arithmetic progression, we have:
\(a_(n+1)=a_n+d, \quad a_(n-1)=a_n-d, \)
where
\(a_n= \frac(a_(n-1) +a_(n+1))(2) \), where \(n>1 \)

Thus, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the two members adjacent to it. This explains the name "arithmetic" progression.

Note that if a 1 and d are given, then the remaining terms of the arithmetic progression can be calculated using the recursive formula a n+1 = a n + d. In this way, it is not difficult to calculate the first few terms of the progression, however, for example, for a 100, a lot of calculations will already be required. Usually, the nth term formula is used for this. According to the definition of an arithmetic progression
\(a_2=a_1+d, \)
\(a_3=a_2+d=a_1+2d, \)
\(a_4=a_3+d=a_1+3d\)
etc.
Generally,
\(a_n=a_1+(n-1)d, \)
since the nth member of an arithmetic progression is obtained from the first member by adding (n-1) times the number d.
This formula is called formula of the nth member of an arithmetic progression.

The sum of the first n terms of an arithmetic progression

Let's find the sum of all natural numbers from 1 to 100.
We write this sum in two ways:
S = l + 2 + 3 + ... + 99 + 100,
S = 100 + 99 + 98 + ... + 2 + 1.
We add these equalities term by term:
2S = 101 + 101 + 101 + ... + 101 + 101.
There are 100 terms in this sum.
Therefore, 2S = 101 * 100, whence S = 101 * 50 = 5050.

Consider now an arbitrary arithmetic progression
a 1 , a 2 , a 3 , ..., a n , ...
Let S n be the sum of the first n terms of this progression:
S n \u003d a 1, a 2, a 3, ..., a n
Then the sum of the first n terms of an arithmetic progression is
\(S_n = n \cdot \frac(a_1+a_n)(2) \)

Since \(a_n=a_1+(n-1)d \), then replacing a n in this formula, we get another formula for finding the sums of the first n terms of an arithmetic progression:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)

Books (textbooks) Abstracts of the Unified State Examination and OGE tests online Games, puzzles Construction of graphs of functions Spelling Dictionary of the Russian Language Dictionary of youth slang Directory of Russian schools Catalog of secondary schools in Russia Catalog of Russian universities List of tasks

For example, the sequence \(2\); \(5\); \(eight\); \(eleven\); \(14\)… is an arithmetic progression, because each next element differs from the previous one by three (can be obtained from the previous one by adding three):

In this progression, the difference \(d\) is positive (equal to \(3\)), and therefore each next term is greater than the previous one. Such progressions are called increasing.

However, \(d\) can also be negative number. For example, in arithmetic progression \(16\); \(ten\); \(four\); \(-2\); \(-8\)… the progression difference \(d\) is equal to minus six.

And in this case, each next element will be less than the previous one. These progressions are called decreasing.

Arithmetic progression notation

Progression is denoted by a small Latin letter.

The numbers that form a progression are called it members(or elements).

They are denoted by the same letter as the arithmetic progression, but with a numerical index equal to the element number in order.

For example, the arithmetic progression \(a_n = \left\( 2; 5; 8; 11; 14…\right\)\) consists of the elements \(a_1=2\); \(a_2=5\); \(a_3=8\) and so on.

In other words, for the progression \(a_n = \left\(2; 5; 8; 11; 14…\right\)\)

Solving problems on an arithmetic progression

In principle, the above information is already enough to solve almost any problem on an arithmetic progression (including those offered at the OGE).

Example (OGE). The arithmetic progression is given by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:

Answer: \(b_5=23\)

Example (OGE). The first three terms of an arithmetic progression are given: \(62; 49; 36…\) Find the value of the first negative term of this progression..
Solution:

	We are given the first elements of the sequence and know that it is an arithmetic progression. That is, each element differs from the neighboring one by the same number. Find out which one by subtracting the previous one from the next element: \(d=49-62=-13\).
	Now we can restore our progression to the desired (first negative) element.
	Ready. You can write an answer.

Answer: \(-3\)

Example (OGE). Several successive elements of an arithmetic progression are given: \(...5; x; 10; 12.5...\) Find the value of the element denoted by the letter \(x\).
Solution:

	To find \(x\), we need to know how much the next element differs from the previous one, in other words, the progression difference. Let's find it from two known neighboring elements: \(d=12.5-10=2.5\).
	And now we find what we are looking for without any problems: \(x=5+2.5=7.5\).
	Ready. You can write an answer.

Answer: \(7,5\).

Example (OGE). Arithmetic progression given following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Solution:

	We need to find the sum of the first six terms of the progression. But we do not know their meanings, we are given only the first element. Therefore, we first calculate the values ​​​​in turn, using the given to us: \(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\) \(n=2\); \(a_(2+1)=a_2+5=-6+5=-1\) \(n=3\); \(a_(3+1)=a_3+5=-1+5=4\) And having calculated the six elements we need, we find their sum.
\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\) \(=(-11)+(-6)+(-1)+4+9+14=9\)	The requested amount has been found.

Answer: \(S_6=9\).

Example (OGE). In arithmetic progression \(a_(12)=23\); \(a_(16)=51\). Find the difference of this progression.
Solution:

Answer: \(d=7\).

Important Arithmetic Progression Formulas

As you can see, many arithmetic progression problems can be solved simply by understanding the main thing - that an arithmetic progression is a chain of numbers, and each next element in this chain is obtained by adding the same number to the previous one (the difference of the progression).

However, sometimes there are situations when it is very inconvenient to solve "on the forehead". For example, imagine that in the very first example, we need to find not the fifth element \(b_5\), but the three hundred and eighty-sixth \(b_(386)\). What is it, we \ (385 \) times to add four? Or imagine that in the penultimate example, you need to find the sum of the first seventy-three elements. Counting is confusing...

Therefore, in such cases, they do not solve “on the forehead”, but use special formulas derived for arithmetic progression. And the main ones are the formula for the nth term of the progression and the formula for the sum \(n\) of the first terms.

Formula for the \(n\)th member: \(a_n=a_1+(n-1)d\), where \(a_1\) is the first member of the progression;
\(n\) – number of the required element;
\(a_n\) is a member of the progression with the number \(n\).

This formula allows us to quickly find at least the three hundredth, even the millionth element, knowing only the first and the progression difference.

Example. The arithmetic progression is given by the conditions: \(b_1=-159\); \(d=8,2\). Find \(b_(246)\).
Solution:

Answer: \(b_(246)=1850\).

The formula for the sum of the first n terms is: \(S_n=\frac(a_1+a_n)(2) \cdot n\), where

\(a_n\) is the last summed term;

Example (OGE). The arithmetic progression is given by the conditions \(a_n=3.4n-0.6\). Find the sum of the first \(25\) terms of this progression.
Solution:

\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\)		To calculate the sum of the first twenty-five elements, we need to know the value of the first and twenty-fifth term. Our progression is given by the formula of the nth term depending on its number (see details). Let's compute the first element by replacing \(n\) with one.
\(n=1;\) \(a_1=3.4 1-0.6=2.8\)		Now let's find the twenty-fifth term by substituting twenty-five instead of \(n\).
\(n=25;\) \(a_(25)=3.4 25-0.6=84.4\)		Well, now we calculate the required amount without any problems.
\(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25=\) \(=\) \(\frac(2,8+84,4)(2)\) \(\cdot 25 =\)\(1090\)		The answer is ready.

Answer: \(S_(25)=1090\).

For the sum \(n\) of the first terms, you can get another formula: you just need to \(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25\ ) instead of \(a_n\) substitute the formula for it \(a_n=a_1+(n-1)d\). We get:

The formula for the sum of the first n terms is: \(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\), where

\(S_n\) – the required sum \(n\) of the first elements;
\(a_1\) is the first term to be summed;
\(d\) – progression difference;
\(n\) - the number of elements in the sum.

Example. Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15,5\); \(fourteen\)…
Solution:

Answer: \(S_(33)=-231\).

More complex arithmetic progression problems

Now you have all the information you need to solve almost any arithmetic progression problem. Let's finish the topic by considering problems in which you need to not only apply formulas, but also think a little (in mathematics, this can be useful ☺)

Example (OGE). Find the sum of all negative terms of the progression: \(-19.3\); \(-19\); \(-18.7\)…
Solution:

\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\)		The task is very similar to the previous one. We start solving the same way: first we find \(d\).
\(d=a_2-a_1=-19-(-19.3)=0.3\)		Now we would substitute \(d\) into the formula for the sum ... and here a small nuance pops up - we don't know \(n\). In other words, we do not know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we get to the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of an arithmetic progression: \(a_n=a_1+(n-1)d\) for our case.
\(a_n=a_1+(n-1)d\)
\(a_n=-19.3+(n-1) 0.3\)		We need \(a_n\) to be greater than zero. Let's find out for what \(n\) this will happen.
\(-19.3+(n-1) 0.3>0\)
\((n-1) 0.3>19.3\) \(|:0.3\)		We divide both sides of the inequality by \(0,3\).
\(n-1>\)\(\frac(19,3)(0,3)\)		We transfer minus one, not forgetting to change signs
\(n>\)\(\frac(19,3)(0,3)\) \(+1\)		Computing...
\(n>65,333…\)		…and it turns out that the first positive element will have the number \(66\). Accordingly, the last negative has \(n=65\). Just in case, let's check it out.
\(n=65;\) \(a_(65)=-19.3+(65-1) 0.3=-0.1\) \(n=66;\) \(a_(66)=-19.3+(66-1) 0.3=0.2\)		Thus, we need to add the first \(65\) elements.
\(S_(65)=\) \(\frac(2 \cdot (-19,3)+(65-1)0,3)(2)\)\(\cdot 65\) \(S_(65)=\)\((-38.6+19.2)(2)\)\(\cdot 65=-630.5\)		The answer is ready.

Answer: \(S_(65)=-630.5\).

Example (OGE). The arithmetic progression is given by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from \(26\)th to \(42\) element inclusive.
Solution:

\(a_1=-33;\) \(a_(n+1)=a_n+4\)	In this problem, you also need to find the sum of elements, but starting not from the first, but from the \(26\)th. We don't have a formula for this. How to decide? Easy - to get the sum from \(26\)th to \(42\)th, you must first find the sum from \(1\)th to \(42\)th, and then subtract from it the sum from the first to \ (25 \) th (see picture).
	For our progression \(a_1=-33\), and the difference \(d=4\) (after all, we add four to the previous element to find the next one). Knowing this, we find the sum of the first \(42\)-uh elements.
\(S_(42)=\) \(\frac(2 \cdot (-33)+(42-1)4)(2)\)\(\cdot 42=\) \(=\)\(\frac(-66+164)(2)\) \(\cdot 42=2058\)	Now the sum of the first \(25\)-th elements.
\(S_(25)=\) \(\frac(2 \cdot (-33)+(25-1)4)(2)\)\(\cdot 25=\) \(=\)\(\frac(-66+96)(2)\) \(\cdot 25=375\)	And finally, we calculate the answer.
\(S=S_(42)-S_(25)=2058-375=1683\)

Answer: \(S=1683\).

For an arithmetic progression, there are several more formulas that we have not considered in this article due to their low practical usefulness. However, you can easily find them.