The formula for the sum of a decreasing arithmetic progression. How to find the sum of an arithmetic progression: formulas and an example of their use

For example, the sequence \(2\); \(5\); \(8\); \(eleven\); \(14\)… is an arithmetic progression, because each next element differs from the previous one by three (can be obtained from the previous one by adding three):

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

However, \(d\) can also be negative number. For example, in arithmetic progression \(16\); \(10\); \(4\); \(-2\); \(-8\)… 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). Arithmetic progression given by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:

Answer: \(b_5=23\)

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

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

Answer: \(-3\)

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

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

Answer: \(7,5\).

Example (OGE). 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.
Solution:

We need to find the sum of the first six terms of the progression. But we do not know their meanings, we are given only the first element. Therefore, we first calculate the values in turn, using the given to us:

\(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\)
\(n=2\); \(a_(2+1)=a_2+5=-6+5=-1\)
\(n=3\); \(a_(3+1)=a_3+5=-1+5=4\)
And having calculated the six elements we need, we find their sum.

\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\)
\(=(-11)+(-6)+(-1)+4+9+14=9\)

The requested amount has been found.

Answer: \(S_6=9\).

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

Answer: \(d=7\).

Important Arithmetic Progression Formulas

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

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

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

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

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

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

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

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

\(a_n\) is the last summed term;

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

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

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

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

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

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

Example. Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15,5\); \(14\)…
Solution:

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

The 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 elementary to quite solid.

First, let's deal with the meaning and formula of the sum. And then we'll decide. For your own pleasure.) The meaning of the sum is as simple as lowing. To find the sum of an arithmetic progression, you just need to carefully add all its members. 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 saves.

The sum formula is simple:

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

S n is the sum of an arithmetic progression. Addition result all members, with first By last. It is important. Add up exactly All members in a row, without gaps and jumps. And, exactly, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of terms five through twentieth, direct application of the formula will be disappointing.)

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

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

n is the 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. Filling question: what kind of member will last, if given endless arithmetic progression?

For a confident answer, you need to understand the elementary meaning of an arithmetic progression and ... read the assignment 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 finite, specific amount just doesn't exist. For the solution, it does not matter what kind of progression is given: finite or infinite. It doesn't matter how it is given: by a series of numbers, or by the formula of the nth member.

The most important thing is to understand that the formula works from the first term of the progression to the term with the 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 the task, all this valuable information is often encrypted, yes ... But nothing, in the examples below we will reveal these secrets.)

Examples of tasks for the sum of an arithmetic progression.

First of all, helpful information:

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

The authors of the assignments 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 just to decipher them. Let's take a 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 the first 10 terms.

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

Where to get the last member number n? Yes, in the same place, in the condition! It says find the sum first 10 members. Well, what number will it be 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, but instead n- ten. Again, the number of the last member is the same as the number of members.

It remains to be determined a 1 And a 10. This is easily calculated by the formula of the nth term, which is given in the problem statement. Don't know how to do it? Visit the previous lesson, without this - nothing.

a 1= 2 1 - 3.5 = -1.5

a 10\u003d 2 10 - 3.5 \u003d 16.5

S n = S 10.

We found out the meaning of all elements of the formula for the sum of an arithmetic progression. It remains to substitute them, and count:

That's all there is to 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 \u003d 2.3. Find the sum of the first 15 terms.

We immediately write the sum formula:

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

a 15 \u003d 2.3 + (15-1) 3.7 \u003d 54.1

It remains to substitute all the elements in 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 just substitute the formula of the nth term, we get:

We give similar ones, we get a new formula for the sum of members of an arithmetic progression:

As you can see, there is no need nth term a n. In some tasks, this formula helps out a lot, yes ... You can remember this formula. And you can simply withdraw it at the right time, as here. After all, the formula for the sum and the formula for the nth term must be remembered in every way.)

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

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

How! No first member, no last, no progression at all... How to live!?

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

Multiples of three... Hm... These are numbers that are evenly 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 a series according to the condition of the problem:

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

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

So, we can safely write down some progression parameters:

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

There are two solutions here. One way is for the super hardworking. You can paint the progression, the whole series of numbers, and count the number of terms with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If the formula is applied to our problem, we get that 99 is the thirtieth member of the progression. Those. n = 30.

We look at the formula for the sum of an arithmetic progression:

We look and rejoice.) We pulled out everything necessary for calculating the amount from the condition of the problem:

a 1= 12.

a 30= 99.

S n = S 30.

What remains is elementary arithmetic. Substitute the numbers in the formula and calculate:

Answer: 1665

Another type of popular puzzles:

4. An arithmetic progression is given:

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

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

We look at the sum formula and ... we are upset.) The formula, let me remind you, calculates the sum 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, paint the entire progression in a row, and put the members from 20 to 34. But ... somehow it turns out stupidly and for a long time, right?)

There is a more elegant solution. Let's break our series into two parts. The first part will from the first term to the nineteenth. Second part - 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 to the sum of the members 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

This shows that to find the sum S 20-34 can be done by simple subtraction

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

Both sums on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Are we getting started?

We extract the progression parameters from the task condition:

d = 1.5.

a 1= -21,5.

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

a 19\u003d -21.5 + (19-1) 1.5 \u003d 5.5

a 34\u003d -21.5 + (34-1) 1.5 \u003d 28

There is nothing left. Subtract the sum of 19 terms from the sum of 34 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 feature in solving this problem. Instead of direct calculation what you need (S 20-34), we counted what, it would seem, is not needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the full result. Such a "feint with the ears" often saves in evil puzzles.)

In this lesson, we examined 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 for the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.

Formula of the nth member:

These formulas will immediately tell you what to look for, in which 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. Arithmetic progression is given by the condition: a 1 =-5.5; a n+1 = a n +0.5. Find the sum of the first 24 terms.

Unusual?) This is a recurrent formula. You can read about it in the previous lesson. Do not ignore the link, such puzzles are often found in the GIA.

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

Difficult?) Will help additional formula from task 2.

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

If you like this site...

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

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

Arithmetic and geometric progressions

Theoretical information

Theoretical information

	Arithmetic progression	Geometric progression
Definition	Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference)	geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression)
Recurrent formula	For any natural n a n + 1 = a n + d	For any natural n b n + 1 = b n ∙ q, b n ≠ 0
nth term formula	a n = a 1 + d (n - 1)	b n \u003d b 1 ∙ q n - 1, b n ≠ 0
characteristic property
Sum of the first n terms

Examples of tasks with comments

Exercise 1

In arithmetic progression ( a n) a 1 = -6, a 2

According to the formula of the nth term:

a 22 = a 1+ d (22 - 1) = a 1+ 21d

By condition:

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

It is necessary to find the difference of progressions:

d= a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = - 48.

Answer : a 22 = -48.

Task 2

Find the fifth term of the geometric progression: -3; 6;....

1st way (using n-term formula)

According to the formula of the n-th member of a geometric progression:

b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.

Because b 1 = -3,

2nd way (using recursive formula)

Since the denominator of the progression is -2 (q = -2), then:

b 3 = 6 ∙ (-2) = -12;

b 4 = -12 ∙ (-2) = 24;

b 5 = 24 ∙ (-2) = -48.

Answer : b 5 = -48.

Task 3

In arithmetic progression ( a n) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.

For an arithmetic progression, the characteristic property has the form .

Therefore:

Substitute the data in the formula:

Answer: 95.

Task 4

In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.

To find the sum of the first n terms of an arithmetic progression, two formulas are used:

Which of them is more convenient to apply in this case?

By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, And a 16 without finding d . Therefore, we use the first formula.

Answer: 368.

Task 5

In arithmetic progression a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.

According to the formula of the nth term:

a 22 = a 1 + d (22 – 1) = a 1+ 21d.

By condition, if a 1= -6, then a 22= -6 + 21d. It is necessary to find the difference of progressions:

d= a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = -48.

Answer : a 22 = -48.

Task 6

Several consecutive terms of a geometric progression are recorded:

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

When solving, we use the formula for the nth term b n \u003d b 1 ∙ q n - 1 For geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.

Substituting the found values into the formula, we get:

Answer : .

Task 7

From the arithmetic progressions given by the formula of the nth term, choose the one for which the condition is satisfied a 27 > 9:

Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:

Answer: 4.

Task 8

In arithmetic progression a 1= 3, d = -1.5. Specify highest value n , for which the inequality a n > -6.

Someone treats the word "progression" with caution, as a very complex term from the branches of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi 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 n-th 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.

Solution. 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

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

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

Progression Difference: Definition

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

d = a(j) - a(j-1).

Allocate:

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 of progression and its arbitrary elements

If 2 arbitrary members of the progression (i-th, k-th) are known, then the difference for this sequence can be established based on the relation:

a(i) = a(k) + (i - k)*d, so d = (a(i) - a(k))/(i-k).

The progression difference and its first term

This expression will help determine the 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 corresponding 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.

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

\(a_1=-33;\) \(a_(n+1)=a_n+4\)	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.
\(S=S_(42)-S_(25)=2058-375=1683\)