Arithmetic progression name a sequence of numbers (terms of a progression)

In which each subsequent term differs from the previous one by a new term, which is also called step or progression difference.

Thus, by specifying the progression step and its first term, you can find any of its elements using the formula

Properties arithmetic progression

1) Each member of an arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next members of the progression

The converse is also true. If the arithmetic mean of adjacent odd (even) terms of a progression is equal to the term that stands between them, then this sequence of numbers is an arithmetic progression. Using this statement, it is very easy to check any sequence.

Also, by the property of arithmetic progression, the above formula can be generalized to the following

This is easy to verify if you write the terms to the right of the equal sign

It is often used in practice to simplify calculations in problems.

2) The sum of the first n terms of an arithmetic progression is calculated using the formula

Remember well the formula for the sum of an arithmetic progression; it is indispensable in calculations and is quite often found in simple life situations.

3) If you need to find not the whole sum, but part of the sequence starting from its kth term, then the following sum formula will be useful to you

4) Of practical interest is finding the sum of n terms of an arithmetic progression starting from the kth number. To do this, use the formula

On this theoretical material ends and we move on to solving common problems in practice.

Example 1. Find the fortieth term of the arithmetic progression 4;7;...

Solution:

According to the condition we have

Let's determine the progression step

Using a well-known formula, we find the fortieth term of the progression

Example 2. An arithmetic progression is given by its third and seventh terms. Find the first term of the progression and the sum of ten.

Solution:

Let us write down the given elements of the progression using the formulas

We subtract the first from the second equation, as a result we find the progression step

We substitute the found value into any of the equations to find the first term of the arithmetic progression

We calculate the sum of the first ten terms of the progression

Without applying complex calculations We found all the required quantities.

Example 3. An arithmetic progression is given by the denominator and one of its terms. Find the first term of the progression, the sum of its 50 terms starting from 50 and the sum of the first 100.

Solution:

Let's write down the formula for the hundredth element of the progression

and find the first one

Based on the first, we find the 50th term of the progression

Finding the sum of the part of the progression

and the sum of the first 100

The progression amount is 250.

Example 4.

Find the number of terms of an arithmetic progression if:

a3-a1=8, a2+a4=14, Sn=111.

Solution:

Let's write the equations in terms of the first term and the progression step and determine them

We substitute the obtained values ​​into the sum formula to determine the number of terms in the sum

We carry out simplifications

and solve the quadratic equation

Of the two values ​​found, only the number 8 fits the problem conditions. Thus, the sum of the first eight terms of the progression is 111.

Example 5.

Solve the equation

1+3+5+...+x=307.

Solution: This equation is the sum of an arithmetic progression. Let's write out its first term and find the difference in progression

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

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

The first element of the sequence;

The fifth element of the sequence;

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

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

The sequence can be set in three ways:

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

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

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

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

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

For example, if , then

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

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

If, for example, , That

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

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

For example, consider the sequence ,

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

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

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

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

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

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

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

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

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

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

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

The main property of an arithmetic progression:

Let's look at the drawing.

We see that

, and at the same time

Adding these two equalities, we get:

.

Let's divide both sides of the equality by 2:

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

Moreover, since

, and at the same time

, That

, and therefore

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

Formula of the th term.

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

and finally

We got formula of the nth term.

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

The sum of n terms of an arithmetic progression.

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

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

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

Let's add in pairs:

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

We get:

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

Let's consider solving arithmetic progression problems.

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

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

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

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

a) Find 31 terms of the progression.

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

A) We see that ;

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

In general

In our case , That's why

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

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

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

The concept of arithmetic progression.

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

I'll write an unfinished series of numbers:

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

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

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

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

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

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

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

First key point.

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

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

Second key point.

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

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

Third key point.

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

That's the whole point.

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

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

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

Terms and designations.

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

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

Arithmetic progression difference.

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

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

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

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

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

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

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

For example:

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

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

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

Arithmetic progression difference usually denoted by the letter d.

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

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

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

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

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

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

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

Let's define d for descending arithmetic progression:

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

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

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

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

Other terms and designations.

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

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

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

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

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

Progressions happen finite and infinite.

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

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

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

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

Or like this, if there are many members:

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

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

(a n), n = 20

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

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

Examples of tasks on arithmetic progression.

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

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

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

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

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

a 3 = a 2 + d

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

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

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

a 4 = a 3 + d

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

a 5 = a 4 + d

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

a 6 = a 5 + d

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

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

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

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

a 1 = a 2 - d

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

That's it. Assignment answer:

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

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

One important conclusion can be drawn from this simple task.

Remember:

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

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

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

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

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

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

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

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

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

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

It remains to write down the answer:

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

Another task:

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

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

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

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

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

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

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

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

Answer: no.

Here's a problem based on real option GIA:

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

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

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

Member numbers? There is not a single number here.

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

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

That's all. Answer: x=12

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

You can.) If you don’t know simple formula, which allows you to solve such tasks in a minute. This formula will be in the next lesson. And this 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. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

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

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

Progression difference: definition

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

d = a(j) – a(j-1).

Highlight:

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

Difference progression and its arbitrary elements

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

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

Difference of progression and its first term

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

Progression difference and its sum

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

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

Sum of an arithmetic progression.

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

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

The formula for the amount is simple:

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

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

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

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

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

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

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

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

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

Examples of tasks on the sum of an arithmetic progression.

First of all, helpful information:

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

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

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

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

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

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

a 1= 2 1 - 3.5 = -1.5

a 10=2·10 - 3.5 =16.5

S n = S 10.

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

That's it. Answer: 75.

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

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

We immediately write the sum formula:

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

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

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

Answer: 423.

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

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

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

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

3. Find the sum of all positive double digit numbers, multiples of three.

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

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

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

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

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

So, we can safely write down some progression parameters:

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

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

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

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

a 1= 12.

a 30= 99.

S n = S 30.

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

Answer: 1665

Another type of popular puzzle:

4. Given an arithmetic progression:

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

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

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

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

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

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

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

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

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

We extract the progression parameters from the problem statement:

d = 1.5.

a 1= -21,5.

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

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

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

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

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

Answer: 262.5

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

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

Practical advice:

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

Formula for the nth term:

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

And now the tasks for independent solution.

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

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

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

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

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

Is it difficult?) The additional formula from task 2 will help.

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

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