The formula for the nth number of arithmetic. How to find an arithmetic progression? Arithmetic progression examples with solution

Someone treats the word "progression" with caution, as a very complex term from the sections of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi counter (where they still remain). And to understand the essence (and in mathematics there is nothing more important than “to understand the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.

Mathematical number sequence

It is customary to call a numerical sequence a series of numbers, each of which has its own number.

and 1 is the first member of the sequence;

and 2 is the second member of the sequence;

and 7 is the seventh member of the sequence;

and n is the nth member of the sequence;

However, not any arbitrary set of figures and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth member is related to its ordinal number by a dependence that can be clearly formulated mathematically. In other words: the numerical value of the nth number is some function of n.

a - value of a member of the numerical sequence;

n is its serial number;

f(n) is a function where the ordinal in the numeric sequence n is the argument.

Definition

An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth member of an arithmetic sequence is as follows:

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

a n+1 - the formula of the next number;

d - difference (a certain number).

It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one, and such an arithmetic progression will be increasing.

In the graph below, it is easy to see why the number sequence is called "increasing".

In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.

The value of the specified member

Sometimes it is necessary to determine the value of some arbitrary term a n of an arithmetic progression. You can do this by calculating successively the values ​​of all members of the arithmetic progression, from the first to the desired one. However, this way is not always acceptable if, for example, it is necessary to find the value of the five thousandth or eight millionth term. The traditional calculation will take a long time. However, a specific arithmetic progression can be investigated using certain formulas. There is also a formula for the nth term: the value of any member of an arithmetic progression can be determined as the sum of the first member of the progression with the difference of the progression, multiplied by the number of the desired member, minus one.

The formula is universal for increasing and decreasing progression.

An example of calculating the value of a given member

Let's solve the following problem of finding the value of the n-th member of an arithmetic progression.

Condition: there is an arithmetic progression with parameters:

The first member of the sequence is 3;

The difference in the number series is 1.2.

Task: it is necessary to find the value of 214 terms

Solution: to determine the value of a given member, we use the formula:

a(n) = a1 + d(n-1)

Substituting the data from the problem statement into the expression, we have:

a(214) = a1 + d(n-1)

a(214) = 3 + 1.2 (214-1) = 258.6

Answer: The 214th member of the sequence is equal to 258.6.

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

Sum of a given number of terms

Very often, in a given arithmetic series, it is required to determine the sum of the values ​​of some of its segments. It also doesn't need to calculate the values ​​of each term and then sum them up. This method is applicable if the number of terms whose sum must be found is small. In other cases, it is more convenient to use the following formula.

The sum of the members of an arithmetic progression from 1 to n is equal to the sum of the first and nth members, multiplied by the member number n and divided by two. If in the formula the value of the n-th member is replaced by the expression from the previous paragraph of the article, we get:

Calculation example

For example, let's solve a problem with the following conditions:

The first term of the sequence is zero;

The difference is 0.5.

In the problem, it is required to determine the sum of the terms of the series from 56 to 101.

Decision. Let's use the formula for determining the sum of the progression:

s(n) = (2∙a1 + d∙(n-1))∙n/2

First, we determine the sum of the values ​​of 101 members of the progression by substituting the given conditions of our problem into the formula:

s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2 525

Obviously, in order to find out the sum of the terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.

s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5

So the sum of the arithmetic progression for this example is:

s 101 - s 55 \u003d 2,525 - 742.5 \u003d 1,782.5

Example of practical application of arithmetic progression

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

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

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

30 - 3 = 27 km.

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

The member number is the number of kilometers traveled (minus the first three).

The value of the member is the sum.

The first term in this problem will be equal to a 1 = 50 rubles.

Progression difference d = 22 p.

the number of interest to us - the value of the (27 + 1)th member of the arithmetic progression - the meter reading at the end of the 27th kilometer - 27.999 ... = 28 km.

a 28 \u003d 50 + 22 ∙ (28 - 1) \u003d 644

Calculations of calendar data for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the luminary. In addition, various numerical series are successfully used in statistics and other applied branches of mathematics.

Another kind of number sequence is geometric

A geometric progression is characterized by a large, compared with an arithmetic, rate of change. It is no coincidence that in politics, sociology, medicine, often, in order to show the high speed of the spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops exponentially.

The N-th member of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first member is 1, the denominator is 2, respectively, then:

n=1: 1 ∙ 2 = 2

n=2: 2 ∙ 2 = 4

n=3: 4 ∙ 2 = 8

n=4: 8 ∙ 2 = 16

n=5: 16 ∙ 2 = 32,

b n - the value of the current member of the geometric progression;

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

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

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

As in the case of arithmetic, a geometric progression has a formula for the value of an arbitrary member. Any n-th term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:

Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Find the 5th term of the progression

b 5 \u003d b 1 ∙ q (5-1) \u003d 3 ∙ 1.5 4 \u003d 15.1875

The sum of a given number of members is also calculated using a special formula. The sum of the first n members of a geometric progression is equal to the difference between the product of the nth member of the progression and its denominator and the first member of the progression, divided by the denominator reduced by one:

If b n is replaced using the formula discussed above, the value of the sum of the first n members of the considered number series will take the form:

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

s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280

Lesson type: learning new material.

Lesson Objectives:

• expansion and deepening of students' ideas about tasks solved using arithmetic progression; organization of search activity of students when deriving the formula for the sum of the first n members of an arithmetic progression;
• development of skills to independently acquire new knowledge, use already acquired knowledge to achieve the task;
• development of the desire and need to generalize the facts obtained, the development of independence.

Tasks:

• generalize and systematize the existing knowledge on the topic “Arithmetic progression”;
• derive formulas for calculating the sum of the first n members of an arithmetic progression;
• teach how to apply the obtained formulas in solving various problems;
• draw students' attention to the procedure for finding the value of a numerical expression.

Equipment:

• cards with tasks for work in groups and pairs;
• evaluation paper;
• presentation"Arithmetic progression".

I. Actualization of basic knowledge.

1. Independent work in pairs.

1st option:

Define an arithmetic progression. Write down a recursive formula that defines an arithmetic progression. Give an example of an arithmetic progression and indicate its difference.

2nd option:

Write down the formula for the nth term of an arithmetic progression. Find the 100th term of an arithmetic progression ( a n}: 2, 5, 8 …
At this time, two students on the back of the board are preparing answers to the same questions.
Students evaluate the partner's work by comparing it with the board. (Leaflets with answers are handed over).

2. Game moment.

Exercise 1.

Teacher. I conceived some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th member of this progression. (1, 3, 5, 7, 9, 11, 13, 15…)

Questions from students.

1. What is the sixth term of the progression and what is the difference?
2. What is the eighth term of the progression and what is the difference?

If there are no more questions, then the teacher can stimulate them - a “ban” on d (difference), that is, it is not allowed to ask what the difference is. You can ask questions: what is the 6th term of the progression and what is the 8th term of the progression?

Task 2.

There are 20 numbers written on the board: 1, 4, 7 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58.

The teacher stands with his back to the blackboard. The students say the number of the number, and the teacher immediately calls the number itself. Explain how I can do it?

The teacher remembers the formula of the nth term a n \u003d 3n - 2 and, substituting the given values ​​of n, finds the corresponding values a n .

II. Statement of the educational task.

I propose to solve an old problem dating back to the 2nd millennium BC, found in Egyptian papyri.

Task:“Let it be said to you: divide 10 measures of barley between 10 people, the difference between each person and his neighbor is 1/8 of the measure.”

• How does this problem relate to the topic of arithmetic progression? (Each next person gets 1/8 of the measure more, so the difference is d=1/8, 10 people, so n=10.)
• What do you think the number 10 means? (The sum of all members of the progression.)
• What else do you need to know to make it easy and simple to divide barley according to the condition of the problem? (The first term of the progression.)

Lesson objective- obtaining the dependence of the sum of the terms of the progression on their number, the first term and the difference, and checking whether the problem was solved correctly in ancient times.

Before deriving the formula, let's see how the ancient Egyptians solved the problem.

And they solved it like this:

1) 10 measures: 10 = 1 measure - average share;
2) 1 measure ∙ = 2 measures - doubled average share.
doubled average the share is the sum of the shares of the 5th and 6th person.
3) 2 measures - 1/8 measure = 1 7/8 measures - twice the share of the fifth person.
4) 1 7/8: 2 = 5/16 - the share of the fifth; and so on, you can find the share of each previous and subsequent person.

We get the sequence:

III. The solution of the task.

1. Work in groups

1st group: Find the sum of 20 consecutive natural numbers: S 20 \u003d (20 + 1) ∙ 10 \u003d 210.

In general

II group: Find the sum of natural numbers from 1 to 100 (Legend of Little Gauss).

S 100 \u003d (1 + 100) ∙ 50 \u003d 5050

Conclusion:

III group: Find the sum of natural numbers from 1 to 21.

Solution: 1+21=2+20=3+19=4+18…

Conclusion:

IV group: Find the sum of natural numbers from 1 to 101.

Conclusion:

This method of solving the considered problems is called the “Gauss method”.

2. Each group presents the solution to the problem on the board.

3. Generalization of the proposed solutions for an arbitrary arithmetic progression:

a 1 , a 2 , a 3 ,…, a n-2 , a n-1 , a n .
S n \u003d a 1 + a 2 + a 3 + a 4 + ... + a n-3 + a n-2 + a n-1 + a n.

We find this sum by arguing similarly:

4. Have we solved the task?(Yes.)

IV. Primary comprehension and application of the obtained formulas in solving problems.

1. Checking the solution of an old problem by the formula.

2. Application of the formula in solving various problems.

3. Exercises for the formation of the ability to apply the formula in solving problems.

A) No. 613

Given :( and n) - arithmetic progression;

(a n): 1, 2, 3, ..., 1500

To find: S 1500

Decision: , and 1 = 1, and 1500 = 1500,

B) Given: ( and n) - arithmetic progression;
(and n): 1, 2, 3, ...
S n = 210

To find: n
Decision:

V. Independent work with mutual verification.

Denis went to work as a courier. In the first month, his salary was 200 rubles, in each subsequent month it increased by 30 rubles. How much did he earn in a year?

Given :( and n) - arithmetic progression;
a 1 = 200, d=30, n=12
To find: S 12
Decision:

Answer: Denis received 4380 rubles for the year.

VI. Homework instruction.

1. p. 4.3 - learn the derivation of the formula.
2. №№ 585, 623 .
3. Compose a problem that would be solved using the formula for the sum of the first n terms of an arithmetic progression.

VII. Summing up the lesson.

1. Score sheet

2. Continue the sentences

• Today in class I learned...
• Learned Formulas...
• I think that …

3. Can you find the sum of numbers from 1 to 500? What method will you use to solve this problem?

Bibliography.

1. Algebra, 9th grade. Textbook for educational institutions. Ed. G.V. Dorofeeva. Moscow: Enlightenment, 2009.

Yes, yes: arithmetic progression is not a toy for you :)

Well, friends, if you are reading this text, then the internal cap evidence tells me that you still do not know what an arithmetic progression is, but you really (no, like this: SOOOOO!) want to know. Therefore, I will not torment you with long introductions and will immediately get down to business.

To start, a couple of examples. Consider several sets of numbers:

• 1; 2; 3; 4; ...
• 15; 20; 25; 30; ...
• $\sqrt(2);\ 2\sqrt(2);\ 3\sqrt(2);...$

What do all these sets have in common? At first glance, nothing. But actually there is something. Namely: each next element differs from the previous one by the same number.

Judge for yourself. The first set is just consecutive numbers, each one more than the previous one. In the second case, the difference between adjacent numbers is already equal to five, but this difference is still constant. In the third case, there are roots in general. However, $2\sqrt(2)=\sqrt(2)+\sqrt(2)$, while $3\sqrt(2)=2\sqrt(2)+\sqrt(2)$, i.e. in which case each next element simply increases by $\sqrt(2)$ (and don't be scared that this number is irrational).

So: all such sequences are just called arithmetic progressions. Let's give a strict definition:

Definition. A sequence of numbers in which each next differs from the previous one by exactly the same amount is called an arithmetic progression. The very amount by which the numbers differ is called the progression difference and is most often denoted by the letter $d$.

Notation: $\left(((a)_(n)) \right)$ is the progression itself, $d$ is its difference.

And just a couple of important remarks. First, progression is considered only orderly sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. You can't rearrange or swap numbers.

Secondly, the sequence itself can be either finite or infinite. For example, the set (1; 2; 3) is obviously a finite arithmetic progression. But if you write something like (1; 2; 3; 4; ...) - this is already an infinite progression. The ellipsis after the four, as it were, hints that quite a lot of numbers go further. Infinitely many, for example. :)

I would also like to note that progressions are increasing and decreasing. We have already seen increasing ones - the same set (1; 2; 3; 4; ...). Here are examples of decreasing progressions:

• 49; 41; 33; 25; 17; ...
• 17,5; 12; 6,5; 1; −4,5; −10; ...
• $\sqrt(5);\ \sqrt(5)-1;\ \sqrt(5)-2;\ \sqrt(5)-3;...$

Okay, okay: the last example may seem overly complicated. But the rest, I think, you understand. Therefore, we introduce new definitions:

Definition. An arithmetic progression is called:

1. increasing if each next element is greater than the previous one;
2. decreasing, if, on the contrary, each subsequent element is less than the previous one.

In addition, there are so-called "stationary" sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).

Only one question remains: how to distinguish an increasing progression from a decreasing one? Fortunately, everything here depends only on the sign of the number $d$, i.e. progression differences:

1. If $d \gt 0$, then the progression is increasing;
2. If $d \lt 0$, then the progression is obviously decreasing;
3. Finally, there is the case $d=0$ — in this case the entire progression is reduced to a stationary sequence of identical numbers: (1; 1; 1; 1; ...), etc.

Let's try to calculate the difference $d$ for the three decreasing progressions above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract from the number on the right, the number on the left. It will look like this:

• 41−49=−8;
• 12−17,5=−5,5;
• $\sqrt(5)-1-\sqrt(5)=-1$.

As you can see, in all three cases the difference really turned out to be negative. And now that we have more or less figured out the definitions, it's time to figure out how progressions are described and what properties they have.

Members of the progression and the recurrent formula

Since the elements of our sequences cannot be interchanged, they can be numbered:

\[\left(((a)_(n)) \right)=\left\( ((a)_(1)),\ ((a)_(2)),((a)_(3 )),... \right\)\]

Individual elements of this set are called members of the progression. They are indicated in this way with the help of a number: the first member, the second member, and so on.

In addition, as we already know, neighboring members of the progression are related by the formula:

\[((a)_(n))-((a)_(n-1))=d\Rightarrow ((a)_(n))=((a)_(n-1))+d \]

In short, to find the $n$th term of the progression, you need to know the $n-1$th term and the difference $d$. Such a formula is called recurrent, because with its help you can find any number, only knowing the previous one (and in fact, all the previous ones). This is very inconvenient, so there is a more tricky formula that reduces any calculation to the first term and the difference:

\[((a)_(n))=((a)_(1))+\left(n-1 \right)d\]

You have probably come across this formula before. They like to give it in all sorts of reference books and reshebniks. And in any sensible textbook on mathematics, it is one of the first.

However, I suggest you practice a little.

Task number 1. Write down the first three terms of the arithmetic progression $\left(((a)_(n)) \right)$ if $((a)_(1))=8,d=-5$.

Decision. So, we know the first term $((a)_(1))=8$ and the progression difference $d=-5$. Let's use the formula just given and substitute $n=1$, $n=2$ and $n=3$:

\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)d; \\ & ((a)_(1))=((a)_(1))+\left(1-1 \right)d=((a)_(1))=8; \\ & ((a)_(2))=((a)_(1))+\left(2-1 \right)d=((a)_(1))+d=8-5= 3; \\ & ((a)_(3))=((a)_(1))+\left(3-1 \right)d=((a)_(1))+2d=8-10= -2. \\ \end(align)\]

Answer: (8; 3; -2)

That's all! Note that our progression is decreasing.

Of course, $n=1$ could not have been substituted - we already know the first term. However, by substituting the unit, we made sure that even for the first term our formula works. In other cases, everything came down to banal arithmetic.

Task number 2. Write out the first three terms of an arithmetic progression if its seventh term is −40 and its seventeenth term is −50.

Decision. We write the condition of the problem in the usual terms:

\[((a)_(7))=-40;\quad ((a)_(17))=-50.\]

\[\left\( \begin(align) & ((a)_(7))=((a)_(1))+6d \\ & ((a)_(17))=((a) _(1))+16d \\ \end(align) \right.\]

\[\left\( \begin(align) & ((a)_(1))+6d=-40 \\ & ((a)_(1))+16d=-50 \\ \end(align) \right.\]

I put the sign of the system because these requirements must be met simultaneously. And now we note that if we subtract the first equation from the second equation (we have the right to do this, because we have a system), we get this:

\[\begin(align) & ((a)_(1))+16d-\left(((a)_(1))+6d \right)=-50-\left(-40 \right); \\ & ((a)_(1))+16d-((a)_(1))-6d=-50+40; \\ & 10d=-10; \\&d=-1. \\ \end(align)\]

Just like that, we found the progression difference! It remains to substitute the found number in any of the equations of the system. For example, in the first:

\[\begin(matrix) ((a)_(1))+6d=-40;\quad d=-1 \\ \Downarrow \\ ((a)_(1))-6=-40; \\ ((a)_(1))=-40+6=-34. \\ \end(matrix)\]

Now, knowing the first term and the difference, it remains to find the second and third terms:

\[\begin(align) & ((a)_(2))=((a)_(1))+d=-34-1=-35; \\ & ((a)_(3))=((a)_(1))+2d=-34-2=-36. \\ \end(align)\]

Ready! Problem solved.

Answer: (-34; -35; -36)

Pay attention to a curious property of the progression that we discovered: if we take the $n$th and $m$th terms and subtract them from each other, then we get the difference of the progression multiplied by the number $n-m$:

\[((a)_(n))-((a)_(m))=d\cdot \left(n-m \right)\]

A simple but very useful property that you should definitely know - with its help, you can significantly speed up the solution of many progression problems. Here is a prime example of this:

Task number 3. The fifth term of the arithmetic progression is 8.4, and its tenth term is 14.4. Find the fifteenth term of this progression.

Decision. Since $((a)_(5))=8.4$, $((a)_(10))=14.4$, and we need to find $((a)_(15))$, we note following:

\[\begin(align) & ((a)_(15))-((a)_(10))=5d; \\ & ((a)_(10))-((a)_(5))=5d. \\ \end(align)\]

But by condition $((a)_(10))-((a)_(5))=14.4-8.4=6$, so $5d=6$, whence we have:

\[\begin(align) & ((a)_(15))-14,4=6; \\ & ((a)_(15))=6+14.4=20.4. \\ \end(align)\]

Answer: 20.4

That's all! We did not need to make any systems of equations and calculate the first term and the difference - everything was decided in just a couple of lines.

Now let's consider another type of problem - the search for negative and positive members of the progression. It is no secret that if the progression increases, while its first term is negative, then sooner or later positive terms will appear in it. And vice versa: the terms of a decreasing progression will sooner or later become negative.

At the same time, it is far from always possible to find this moment “on the forehead”, sequentially sorting through the elements. Often, tasks are designed in such a way that without knowing the formulas, calculations would take several sheets - we would simply fall asleep until we found the answer. Therefore, we will try to solve these problems in a faster way.

Task number 4. How many negative terms in an arithmetic progression -38.5; -35.8; …?

Decision. So, $((a)_(1))=-38.5$, $((a)_(2))=-35.8$, from which we immediately find the difference:

Note that the difference is positive, so the progression is increasing. The first term is negative, so indeed at some point we will stumble upon positive numbers. The only question is when this will happen.

Let's try to find out: how long (i.e., up to what natural number $n$) the negativity of the terms is preserved:

\[\begin(align) & ((a)_(n)) \lt 0\Rightarrow ((a)_(1))+\left(n-1 \right)d \lt 0; \\ & -38.5+\left(n-1 \right)\cdot 2.7 \lt 0;\quad \left| \cdot 10 \right. \\ & -385+27\cdot \left(n-1 \right) \lt 0; \\ & -385+27n-27 \lt 0; \\ & 27n \lt 412; \\ & n \lt 15\frac(7)(27)\Rightarrow ((n)_(\max ))=15. \\ \end(align)\]

The last line needs clarification. So we know that $n \lt 15\frac(7)(27)$. On the other hand, only integer values ​​of the number will suit us (moreover: $n\in \mathbb(N)$), so the largest allowable number is precisely $n=15$, and in no case 16.

Task number 5. In arithmetic progression $(()_(5))=-150,(()_(6))=-147$. Find the number of the first positive term of this progression.

This would be exactly the same problem as the previous one, but we don't know $((a)_(1))$. But the neighboring terms are known: $((a)_(5))$ and $((a)_(6))$, so we can easily find the progression difference:

In addition, let's try to express the fifth term in terms of the first and the difference using the standard formula:

\[\begin(align) & ((a)_(n))=((a)_(1))+\left(n-1 \right)\cdot d; \\ & ((a)_(5))=((a)_(1))+4d; \\ & -150=((a)_(1))+4\cdot 3; \\ & ((a)_(1))=-150-12=-162. \\ \end(align)\]

Now we proceed by analogy with the previous problem. We find out at what point in our sequence positive numbers will appear:

\[\begin(align) & ((a)_(n))=-162+\left(n-1 \right)\cdot 3 \gt 0; \\ & -162+3n-3 \gt 0; \\ & 3n \gt 165; \\ & n \gt 55\Rightarrow ((n)_(\min ))=56. \\ \end(align)\]

The minimum integer solution of this inequality is the number 56.

Please note that in the last task everything was reduced to strict inequality, so the option $n=55$ will not suit us.

Now that we have learned how to solve simple problems, let's move on to more complex ones. But first, let's learn another very useful property of arithmetic progressions, which will save us a lot of time and unequal cells in the future. :)

Arithmetic mean and equal indents

Consider several consecutive terms of the increasing arithmetic progression $\left(((a)_(n)) \right)$. Let's try to mark them on a number line:

Arithmetic progression members on the number line

I specifically noted the arbitrary members $((a)_(n-3)),...,((a)_(n+3))$, and not any $((a)_(1)) ,\ ((a)_(2)),\ ((a)_(3))$ etc. Because the rule, which I will now tell you, works the same for any "segments".

And the rule is very simple. Let's remember the recursive formula and write it down for all marked members:

\[\begin(align) & ((a)_(n-2))=((a)_(n-3))+d; \\ & ((a)_(n-1))=((a)_(n-2))+d; \\ & ((a)_(n))=((a)_(n-1))+d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n+1))+d; \\ \end(align)\]

However, these equalities can be rewritten differently:

\[\begin(align) & ((a)_(n-1))=((a)_(n))-d; \\ & ((a)_(n-2))=((a)_(n))-2d; \\ & ((a)_(n-3))=((a)_(n))-3d; \\ & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(n+3))=((a)_(n))+3d; \\ \end(align)\]

Well, so what? But the fact that the terms $((a)_(n-1))$ and $((a)_(n+1))$ lie at the same distance from $((a)_(n)) $. And this distance is equal to $d$. The same can be said about the terms $((a)_(n-2))$ and $((a)_(n+2))$ - they are also removed from $((a)_(n))$ by the same distance equal to $2d$. You can continue indefinitely, but the picture illustrates the meaning well

The members of the progression lie at the same distance from the center

What does this mean for us? This means that you can find $((a)_(n))$ if the neighboring numbers are known:

\[((a)_(n))=\frac(((a)_(n-1))+((a)_(n+1)))(2)\]

We have deduced a magnificent statement: each member of an arithmetic progression is equal to the arithmetic mean of the neighboring members! Moreover, we can deviate from our $((a)_(n))$ to the left and to the right not by one step, but by $k$ steps — and still the formula will be correct:

\[((a)_(n))=\frac(((a)_(n-k))+((a)_(n+k)))(2)\]

Those. we can easily find some $((a)_(150))$ if we know $((a)_(100))$ and $((a)_(200))$, because $(( a)_(150))=\frac(((a)_(100))+((a)_(200)))(2)$. At first glance, it may seem that this fact does not give us anything useful. However, in practice, many tasks are specially "sharpened" for the use of the arithmetic mean. Take a look:

Task number 6. Find all values ​​of $x$ such that the numbers $-6((x)^(2))$, $x+1$ and $14+4((x)^(2))$ are consecutive members of an arithmetic progression (in specified order).

Decision. Since these numbers are members of a progression, the arithmetic mean condition is satisfied for them: the central element $x+1$ can be expressed in terms of neighboring elements:

\[\begin(align) & x+1=\frac(-6((x)^(2))+14+4((x)^(2)))(2); \\ & x+1=\frac(14-2((x)^(2)))(2); \\ & x+1=7-((x)^(2)); \\ & ((x)^(2))+x-6=0. \\ \end(align)\]

The result is a classic quadratic equation. Its roots: $x=2$ and $x=-3$ are the answers.

Answer: -3; 2.

Task number 7. Find the values ​​of $$ such that the numbers $-1;4-3;(()^(2))+1$ form an arithmetic progression (in that order).

Decision. Again, we express the middle term in terms of the arithmetic mean of neighboring terms:

\[\begin(align) & 4x-3=\frac(x-1+((x)^(2))+1)(2); \\ & 4x-3=\frac(((x)^(2))+x)(2);\quad \left| \cdot 2\right.; \\ & 8x-6=((x)^(2))+x; \\ & ((x)^(2))-7x+6=0. \\ \end(align)\]

Another quadratic equation. And again two roots: $x=6$ and $x=1$.

Answer: 1; 6.

If in the process of solving a problem you get some brutal numbers, or you are not completely sure of the correctness of the answers found, then there is a wonderful trick that allows you to check: did we solve the problem correctly?

Let's say in problem 6 we got answers -3 and 2. How can we check that these answers are correct? Let's just plug them into the original condition and see what happens. Let me remind you that we have three numbers ($-6(()^(2))$, $+1$ and $14+4(()^(2))$), which should form an arithmetic progression. Substitute $x=-3$:

\[\begin(align) & x=-3\Rightarrow \\ & -6((x)^(2))=-54; \\ &x+1=-2; \\ & 14+4((x)^(2))=50. \end(align)\]

We got the numbers -54; −2; 50 that differ by 52 is undoubtedly an arithmetic progression. The same thing happens for $x=2$:

\[\begin(align) & x=2\Rightarrow \\ & -6((x)^(2))=-24; \\ &x+1=3; \\ & 14+4((x)^(2))=30. \end(align)\]

Again a progression, but with a difference of 27. Thus, the problem is solved correctly. Those who wish can check the second task on their own, but I’ll say right away: everything is correct there too.

In general, while solving the last problems, we stumbled upon another interesting fact that also needs to be remembered:

If three numbers are such that the second is the average of the first and last, then these numbers form an arithmetic progression.

In the future, understanding this statement will allow us to literally “construct” the necessary progressions based on the condition of the problem. But before we engage in such a "construction", we should pay attention to one more fact, which directly follows from what has already been considered.

Grouping and sum of elements

Let's go back to the number line again. We note there several members of the progression, between which, perhaps. worth a lot of other members:

6 elements marked on the number line

Let's try to express the "left tail" in terms of $((a)_(n))$ and $d$, and the "right tail" in terms of $((a)_(k))$ and $d$. It's very simple:

\[\begin(align) & ((a)_(n+1))=((a)_(n))+d; \\ & ((a)_(n+2))=((a)_(n))+2d; \\ & ((a)_(k-1))=((a)_(k))-d; \\ & ((a)_(k-2))=((a)_(k))-2d. \\ \end(align)\]

Now note that the following sums are equal:

\[\begin(align) & ((a)_(n))+((a)_(k))=S; \\ & ((a)_(n+1))+((a)_(k-1))=((a)_(n))+d+((a)_(k))-d= S; \\ & ((a)_(n+2))+((a)_(k-2))=((a)_(n))+2d+((a)_(k))-2d= S. \end(align)\]

Simply put, if we consider as a start two elements of the progression, which in total are equal to some number $S$, and then we start stepping from these elements in opposite directions (towards each other or vice versa to move away), then the sums of the elements that we will stumble upon will also be equal$S$. This can be best represented graphically:

Same indents give equal sums

Understanding this fact will allow us to solve problems of a fundamentally higher level of complexity than those that we considered above. For example, these:

Task number 8. Determine the difference of an arithmetic progression in which the first term is 66, and the product of the second and twelfth terms is the smallest possible.

Decision. Let's write down everything we know:

\[\begin(align) & ((a)_(1))=66; \\&d=? \\ & ((a)_(2))\cdot ((a)_(12))=\min . \end(align)\]

So, we do not know the difference of the progression $d$. Actually, the whole solution will be built around the difference, since the product $((a)_(2))\cdot ((a)_(12))$ can be rewritten as follows:

\[\begin(align) & ((a)_(2))=((a)_(1))+d=66+d; \\ & ((a)_(12))=((a)_(1))+11d=66+11d; \\ & ((a)_(2))\cdot ((a)_(12))=\left(66+d \right)\cdot \left(66+11d \right)= \\ & =11 \cdot \left(d+66 \right)\cdot \left(d+6 \right). \end(align)\]

For those in the tank: I've taken the common factor 11 out of the second bracket. Thus, the desired product is a quadratic function with respect to the variable $d$. Therefore, consider the function $f\left(d \right)=11\left(d+66 \right)\left(d+6 \right)$ - its graph will be a parabola with branches up, because if we open the brackets, we get:

\[\begin(align) & f\left(d \right)=11\left(((d)^(2))+66d+6d+66\cdot 6 \right)= \\ & =11(( d)^(2))+11\cdot 72d+11\cdot 66\cdot 6 \end(align)\]

As you can see, the coefficient with the highest term is 11 - this is a positive number, so we are really dealing with a parabola with branches up:

graph of a quadratic function - parabola

Please note: this parabola takes its minimum value at its vertex with the abscissa $((d)_(0))$. Of course, we can calculate this abscissa according to the standard scheme (there is a formula $((d)_(0))=(-b)/(2a)\;$), but it would be much more reasonable to note that the desired vertex lies on the axis symmetry of the parabola, so the point $((d)_(0))$ is equidistant from the roots of the equation $f\left(d \right)=0$:

\[\begin(align) & f\left(d\right)=0; \\ & 11\cdot \left(d+66 \right)\cdot \left(d+6 \right)=0; \\ & ((d)_(1))=-66;\quad ((d)_(2))=-6. \\ \end(align)\]

That is why I was in no hurry to open the brackets: in the original form, the roots were very, very easy to find. Therefore, the abscissa is equal to the arithmetic mean of the numbers −66 and −6:

\[((d)_(0))=\frac(-66-6)(2)=-36\]

What gives us the discovered number? With it, the required product takes the smallest value (by the way, we did not calculate $((y)_(\min ))$ - this is not required of us). At the same time, this number is the difference of the initial progression, i.e. we found the answer. :)

Answer: -36

Task number 9. Insert three numbers between the numbers $-\frac(1)(2)$ and $-\frac(1)(6)$ so that together with the given numbers they form an arithmetic progression.

Decision. In fact, we need to make a sequence of five numbers, with the first and last number already known. Denote the missing numbers by the variables $x$, $y$ and $z$:

\[\left(((a)_(n)) \right)=\left\( -\frac(1)(2);x;y;z;-\frac(1)(6) \right\ )\]

Note that the number $y$ is the "middle" of our sequence - it is equidistant from the numbers $x$ and $z$, and from the numbers $-\frac(1)(2)$ and $-\frac(1)( 6)$. And if at the moment we cannot get $y$ from the numbers $x$ and $z$, then the situation is different with the ends of the progression. Remember the arithmetic mean:

Now, knowing $y$, we will find the remaining numbers. Note that $x$ lies between $-\frac(1)(2)$ and $y=-\frac(1)(3)$ just found. So

Arguing similarly, we find the remaining number:

Ready! We found all three numbers. Let's write them down in the answer in the order in which they should be inserted between the original numbers.

Answer: $-\frac(5)(12);\ -\frac(1)(3);\ -\frac(1)(4)$

Task number 10. Between the numbers 2 and 42, insert several numbers that, together with the given numbers, form an arithmetic progression, if it is known that the sum of the first, second, and last of the inserted numbers is 56.

Decision. An even more difficult task, which, however, is solved in the same way as the previous ones - through the arithmetic mean. The problem is that we don't know exactly how many numbers to insert. Therefore, for definiteness, we assume that after inserting there will be exactly $n$ numbers, and the first of them is 2, and the last is 42. In this case, the desired arithmetic progression can be represented as:

\[\left(((a)_(n)) \right)=\left\( 2;((a)_(2));((a)_(3));...;(( a)_(n-1));42 \right\)\]

\[((a)_(2))+((a)_(3))+((a)_(n-1))=56\]

Note, however, that the numbers $((a)_(2))$ and $((a)_(n-1))$ are obtained from the numbers 2 and 42 standing at the edges by one step towards each other, i.e. . to the center of the sequence. And this means that

\[((a)_(2))+((a)_(n-1))=2+42=44\]

But then the above expression can be rewritten like this:

\[\begin(align) & ((a)_(2))+((a)_(3))+((a)_(n-1))=56; \\ & \left(((a)_(2))+((a)_(n-1)) \right)+((a)_(3))=56; \\ & 44+((a)_(3))=56; \\ & ((a)_(3))=56-44=12. \\ \end(align)\]

Knowing $((a)_(3))$ and $((a)_(1))$, we can easily find the progression difference:

\[\begin(align) & ((a)_(3))-((a)_(1))=12-2=10; \\ & ((a)_(3))-((a)_(1))=\left(3-1 \right)\cdot d=2d; \\ & 2d=10\Rightarrow d=5. \\ \end(align)\]

It remains only to find the remaining members:

\[\begin(align) & ((a)_(1))=2; \\ & ((a)_(2))=2+5=7; \\ & ((a)_(3))=12; \\ & ((a)_(4))=2+3\cdot 5=17; \\ & ((a)_(5))=2+4\cdot 5=22; \\ & ((a)_(6))=2+5\cdot 5=27; \\ & ((a)_(7))=2+6\cdot 5=32; \\ & ((a)_(8))=2+7\cdot 5=37; \\ & ((a)_(9))=2+8\cdot 5=42; \\ \end(align)\]

Thus, already at the 9th step we will come to the left end of the sequence - the number 42. In total, only 7 numbers had to be inserted: 7; 12; 17; 22; 27; 32; 37.

Answer: 7; 12; 17; 22; 27; 32; 37

Text tasks with progressions

In conclusion, I would like to consider a couple of relatively simple problems. Well, as simple ones: for most students who study mathematics at school and have not read what is written above, these tasks may seem like a gesture. Nevertheless, it is precisely such tasks that come across in the OGE and the USE in mathematics, so I recommend that you familiarize yourself with them.

Task number 11. The team produced 62 parts in January, and in each subsequent month they produced 14 more parts than in the previous one. How many parts did the brigade produce in November?

Decision. Obviously, the number of parts, painted by month, will be an increasing arithmetic progression. And:

\[\begin(align) & ((a)_(1))=62;\quad d=14; \\ & ((a)_(n))=62+\left(n-1 \right)\cdot 14. \\ \end(align)\]

November is the 11th month of the year, so we need to find $((a)_(11))$:

\[((a)_(11))=62+10\cdot 14=202\]

Therefore, 202 parts will be manufactured in November.

Task number 12. The bookbinding workshop bound 216 books in January, and each month it bound 4 more books than the previous month. How many books did the workshop bind in December?

Decision. All the same:

$\begin(align) & ((a)_(1))=216;\quad d=4; \\ & ((a)_(n))=216+\left(n-1 \right)\cdot 4. \\ \end(align)$

December is the last, 12th month of the year, so we are looking for $((a)_(12))$:

\[((a)_(12))=216+11\cdot 4=260\]

This is the answer - 260 books will be bound in December.

Well, if you have read this far, I hasten to congratulate you: you have successfully completed the “young fighter course” in arithmetic progressions. We can safely move on to the next lesson, where we will study the progression sum formula, as well as important and very useful consequences from it.

Mathematics has its own beauty, as does painting and poetry.

Russian scientist, mechanic N.E. Zhukovsky

Very common tasks in the entrance tests in mathematics are tasks related to the concept of an arithmetic progression. To successfully solve such problems, it is necessary to know the properties of an arithmetic progression well and have certain skills in their application.

Let us first recall the main properties of an arithmetic progression and present the most important formulas, associated with this concept.

Definition. Numeric sequence, in which each subsequent term differs from the previous one by the same number, called an arithmetic progression. At the same time, the numberis called the progression difference.

For an arithmetic progression, the formulas are valid

, (1)

where . Formula (1) is called the formula of the common term of an arithmetic progression, and formula (2) is the main property of an arithmetic progression: each member of the progression coincides with the arithmetic mean of its neighboring members and .

Note that it is precisely because of this property that the progression under consideration is called "arithmetic".

Formulas (1) and (2) above are summarized as follows:

(3)

To calculate the sum first members of an arithmetic progressionthe formula is usually used

(5) where and .

If we take into account the formula (1), then formula (5) implies

If we designate

where . Since , then formulas (7) and (8) are a generalization of the corresponding formulas (5) and (6).

In particular , from formula (5) it follows, what

Among the little-known to most students is the property of an arithmetic progression, formulated by means of the following theorem.

Theorem. If , then

Proof. If , then

The theorem has been proven.

For example , using the theorem, it can be shown that

Let's move on to the consideration of typical examples of solving problems on the topic "Arithmetic progression".

Example 1 Let and . To find .

Decision. Applying formula (6), we obtain . Since and , then or .

Example 2 Let three times more, and when dividing by in the quotient, it turns out 2 and the remainder is 8. Determine and.

Decision. The system of equations follows from the condition of the example

Since , , and , then from the system of equations (10) we obtain

The solution of this system of equations are and .

Example 3 Find if and .

Decision. According to formula (5), we have or . However, using property (9), we obtain .

Since and , then from the equality the equation follows or .

Example 4 Find if .

Decision.By formula (5) we have

However, using the theorem, one can write

From here and from formula (11) we obtain .

Example 5. Given: . To find .

Decision. Since , then . However , therefore .

Example 6 Let , and . To find .

Decision. Using formula (9), we obtain . Therefore, if , then or .

Since and then here we have a system of equations

Solving which, we get and .

Natural root of the equation is an .

Example 7 Find if and .

Decision. Since according to formula (3) we have that , then the system of equations follows from the condition of the problem

If we substitute the expressioninto the second equation of the system, then we get or .

The roots of the quadratic equation are and .

Let's consider two cases.

1. Let , then . Since and , then .

In this case, according to formula (6), we have

2. If , then , and

Answer: and.

Example 8 It is known that and To find .

Decision. Taking into account formula (5) and the condition of the example, we write and .

This implies the system of equations

If we multiply the first equation of the system by 2, and then add it to the second equation, we get

According to formula (9), we have. In this connection, from (12) it follows or .

Since and , then .

Answer: .

Example 9 Find if and .

Decision. Since , and by condition , then or .

From formula (5) it is known, what . Since , then .

Hence , here we have a system of linear equations

From here we get and . Taking into account formula (8), we write .

Example 10 Solve the equation.

Decision. It follows from the given equation that . Let's assume that , , and . In this case .

According to formula (1), we can write or .

Since , equation (13) has a unique suitable root .

Example 11. Find the maximum value provided that and .

Decision. Since , then the considered arithmetic progression is decreasing. In this regard, the expression takes on a maximum value when it is the number of the minimum positive member of the progression.

We use formula (1) and the fact, which and . Then we get that or .

Because , then or . However, in this inequalitylargest natural number, That's why .

If the values ​​, and are substituted into formula (6), then we get .

Answer: .

Example 12. Find the sum of all two-digit natural numbers that, when divided by 6, have a remainder of 5.

Decision. Denote by the set of all two-valued natural numbers, i.e. . Next, we construct a subset consisting of those elements (numbers) of the set that, when divided by the number 6, give a remainder of 5.

Easy to install, what . Obviously , that the elements of the setform an arithmetic progression, in which and .

To determine the cardinality (number of elements) of the set, we assume that . Since and , then formula (1) implies or . Taking into account formula (5), we obtain .

The above examples of solving problems can by no means claim to be exhaustive. This article is written on the basis of an analysis of modern methods for solving typical problems on a given topic. For a deeper study of methods for solving problems related to arithmetic progression, it is advisable to refer to the list of recommended literature.

1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.

2. Suprun V.P. Mathematics for high school students: additional sections of the school curriculum. – M.: Lenand / URSS, 2014. - 216 p.

3. Medynsky M.M. A complete course of elementary mathematics in tasks and exercises. Book 2: Number Sequences and Progressions. – M.: Editus, 2015. - 208 p.

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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 a negative number. for example, in arithmetic progression \(16\); \(ten\); \(4\); \(-2\); \(-8\)… 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 \(b_5\).
Decision:

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..
Decision:

	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 \(x\).
Decision:

	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). The arithmetic progression is given by the following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Decision:

	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.
Decision:

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 \(b_(246)\).
Decision:

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.
Decision:

\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\)		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: \(17\); \(15,5\); \(fourteen\)…
Decision:

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: \(-19.3\); \(-nineteen\); \(-18.7\)…
Decision:

\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\)		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.
Decision:

\(a_1=-33;\) \(a_(n+1)=a_n+4\)	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.