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 of an 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

This concludes the theoretical material and moves 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 using 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 decide 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

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 do not yet know what an arithmetic progression is, but you really (no, like that: SOOOOO!) want to know. Therefore, I will not torment you with long introductions and will get straight to the point.

First, a couple of examples. Let's look at 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 simply consecutive numbers, each next being one more than the previous one. In the second case, the difference between the series standing numbers is already equal to five, but this difference is still constant. In the third case, there are roots altogether. However, $2\sqrt(2)=\sqrt(2)+\sqrt(2)$, and $3\sqrt(2)=2\sqrt(2)+\sqrt(2)$, i.e. and in this case, each next element simply increases by $\sqrt(2)$ (and don’t be afraid that this number is irrational).

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

Definition. A sequence of numbers in which each next one 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 notes. Firstly, progression is only considered ordered sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. Numbers cannot be rearranged or swapped.

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 in the spirit (1; 2; 3; 4; ...) - this is already an infinite progression. The ellipsis after the four seems to hint that there are quite a few more numbers to come. Infinitely many, for example. :)

I would also like to note that progressions can be increasing or 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 increases;
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 given above. To do this, it is enough to take any two adjacent elements (for example, the first and second) and subtract the number on the left from the number on the right. It will look like this:

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

As we can see, in all three cases the difference actually 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.

Progression terms and recurrence formula

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

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

The individual elements of this set are called members of a progression. They are indicated by a number: first member, second member, etc.

In addition, as we already know, neighboring terms 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 a progression, you need to know the $n-1$th term and the difference $d$. This formula is called recurrent, because with its help you can find any number only by knowing the previous one (and in fact, all the previous ones). This is very inconvenient, so there is a more cunning formula that reduces any calculations to the first term and the difference:

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

You've probably already come across this formula. They like to give it in all sorts of reference books and solution books. And in any sensible mathematics textbook it is one of the first.

However, I suggest you practice a little.

Task No. 1. Write down the first three terms of the arithmetic progression $\left(((a)_(n)) \right)$ if $((a)_(1))=8,d=-5$.

Solution. So, we know the first term $((a)_(1))=8$ and the difference of the progression $d=-5$. 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! Please note: our progression is decreasing.

Of course, $n=1$ could not be substituted - the first term is already known to us. However, by substituting unity, we were convinced that even for the first term our formula works. In other cases, everything came down to banal arithmetic.

Task No. 2. Write down the first three terms of an arithmetic progression if its seventh term is equal to −40 and its seventeenth term is equal to −50.

Solution. Let's write the problem condition in familiar 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 system sign because these requirements must be met simultaneously. Now let’s note that if we subtract the first from the second equation (we have the right to do this, since we have a system), we get this:

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

That's how easy it is to find the progression difference! All that remains is to substitute the found number into 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! The problem is solved.

Answer: (−34; −35; −36)

Notice the interesting property of progression that we discovered: if we take the $n$th and $m$th terms and subtract them from each other, we get the difference of the progression multiplied by the $n-m$ number:

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

Simple but very useful property, which you definitely need to know - with its help you can significantly speed up the solution of many progression problems. Here is a clear example of this:

Task No. 3. The fifth term of an arithmetic progression is 8.4, and its tenth term is 14.4. Find the fifteenth term of this progression.

Solution. 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 $((a)_(10))-((a)_(5))=14.4-8.4=6$, therefore $5d=6$, from which 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 didn’t need to create any systems of equations and calculate the first term and the difference - everything was solved in just a couple of lines.

Now let's look at another type of problem - searching for negative and positive terms of a progression. It is no secret that if a progression increases, and 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 not always possible to find this moment “head-on” by sequentially going through the elements. Often, problems are written in such a way that without knowing the formulas, the calculations would take several sheets of paper—we would simply fall asleep while we found the answer. Therefore, let's try to solve these problems in a faster way.

Task No. 4. How many negative terms are there in the arithmetic progression −38.5; −35.8; ...?

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

Note that the difference is positive, so the progression increases. 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 remains:

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

The last line requires some explanation. So we know that $n \lt 15\frac(7)(27)$. On the other hand, we are satisfied with only integer values ​​of the number (moreover: $n\in \mathbb(N)$), so the largest permissible number is precisely $n=15$, and in no case 16.

Task No. 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 do not know $((a)_(1))$. But the neighboring terms are known: $((a)_(5))$ and $((a)_(6))$, so we can easily find the difference of the progression:

In addition, let's try to express the fifth term through 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 task. Let's 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 to this inequality is the number 56.

Please note: in the last task everything came down 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 study 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 indentations

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

Terms of an arithmetic progression on the number line

I specifically marked arbitrary terms $((a)_(n-3)),...,((a)_(n+3))$, and not some $((a)_(1)) ,\ ((a)_(2)),\ ((a)_(3))$, etc. Because the rule that I’ll tell you about now works the same for any “segments”.

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

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

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? And 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 $((a)_(n))$ at the same distance equal to $2d$. We can continue ad infinitum, but the meaning is well illustrated by the picture

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

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

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

We have derived an excellent statement: every term of an arithmetic progression is equal to the arithmetic mean of its neighboring terms! Moreover: we can step back from our $((a)_(n))$ to the left and to the right not by one step, but by $k$ steps - and the formula will still 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 problems are specially tailored to use the arithmetic mean. Take a look:

Task No. 6. Find all values ​​of $x$ for which the numbers $-6((x)^(2))$, $x+1$ and $14+4((x)^(2))$ are consecutive terms of an arithmetic progression (in in the order indicated).

Solution. Because the specified numbers are members of a progression, for them the condition of the arithmetic mean is satisfied: 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 No. 7. Find the values ​​of $$ for which the numbers $-1;4-3;(()^(2))+1$ form an arithmetic progression (in that order).

Solution. Let's express again average member through the arithmetic mean of neighboring terms:

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

Quadratic equation again. And again there are two roots: $x=6$ and $x=1$.

Answer: 1; 6.

If in the process of solving a problem you come up with some brutal numbers, or you are not entirely sure of the correctness of the answers found, then there is a wonderful technique that allows you to check: have we solved the problem correctly?

Let's say in problem No. 6 we received 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 must form an arithmetic progression. Let's 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 was solved correctly. Those who wish can check the second problem on their own, but I’ll say right away: everything is correct there too.

In general, while solving the last problems, we came across another interesting fact, which also needs to be remembered:

If three numbers are such that the second is the arithmetic mean 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 conditions of the problem. But before we engage in such “construction”, we should pay attention to one more fact, which directly follows from what has already been discussed.

Grouping and summing elements

Let's return to the number axis again. Let us note there several members of the progression, between which, perhaps. is worth a lot of other members:

There are 6 elements marked on the number line

Let's try to express the “left tail” through $((a)_(n))$ and $d$, and the “right tail” through $((a)_(k))$ 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 amounts 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 begin to step from these elements in opposite directions (toward 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 most clearly represented graphically:

Equal indentations give equal amounts

Understanding this fact will allow us to solve problems in a fundamentally more high level difficulties than those we considered above. For example, these:

Task No. 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.

Solution. 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 progression difference $d$. Actually, the entire 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 took the total multiplier of 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 expand 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 of the highest term is 11 - this is a positive number, so we are really dealing with a parabola with upward branches:

schedule 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 by standard scheme(there is the formula $((d)_(0))=(-b)/(2a)\;$), but it would be much more reasonable to note that the desired vertex lies on the axis of 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 particular hurry to open the brackets: in their original form, the roots were very, very easy to find. Therefore, the abscissa is equal to the mean arithmetic numbers−66 and −6:

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

What does the discovered number give us? With it, the required product takes on the smallest value (by the way, we never calculated $((y)_(\min ))$ - this is not required of us). At the same time, this number is the difference from the original progression, i.e. we found the answer. :)

Answer: −36

Task No. 9. Between the numbers $-\frac(1)(2)$ and $-\frac(1)(6)$ insert three numbers so that together with these numbers they form an arithmetic progression.

Solution. Essentially, we need to make a sequence of five numbers, with the first and last number already known. Let's 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 from the numbers $x$ and $z$ we are in this moment we can’t get $y$, then the situation is different with the ends of the progression. Let's remember the arithmetic mean:

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

Using similar reasoning, we find the remaining number:

Ready! We found all three numbers. Let's write them 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 No. 10. Between the numbers 2 and 42, insert several numbers that, together with these numbers, form an arithmetic progression, if you know that the sum of the first, second and last of the inserted numbers is 56.

Solution. Even more difficult task, which, however, is solved according to the same scheme as the previous ones - through the arithmetic mean. The problem is that we don’t know exactly how many numbers need to be inserted. Therefore, let us assume for definiteness that after inserting everything there will be exactly $n$ numbers, and the first of them is 2, and the last is 42. In this case, the required arithmetic progression can be represented in the form:

\[\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 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 expression written above can be rewritten as follows:

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

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

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

All that remains is to find the remaining terms:

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

Thus, already at the 9th step we will arrive at 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

Word problems with progressions

In conclusion, I would like to consider a couple of relatively simple problems. Well, as simple as that: for most students who study mathematics at school and have not read what is written above, these problems may seem tough. Nevertheless, these are the types of problems that appear in the OGE and the Unified State Exam in mathematics, so I recommend that you familiarize yourself with them.

Task No. 11. The team produced 62 parts in January, and in each subsequent month they produced 14 more parts than in the previous month. How many parts did the team produce in November?

Solution. Obviously, the number of parts listed by month will represent an increasing arithmetic progression. Moreover:

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

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 produced in November.

Task No. 12. The bookbinding workshop bound 216 books in January, and in each subsequent month it bound 4 more books than in the previous month. How many books did the workshop bind in December?

Solution. 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’s course” in arithmetic progressions. You can safely move on to the next lesson, where we will study the formula for the sum of progression, as well as important and very useful consequences from it.

Online calculator.
Solving an arithmetic progression.
Given: a n , d, n
Find: a 1

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

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

This online calculator may be useful for high school students in preparing for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.

This way you can spend your own training and/or training their younger brothers or sisters, while the level of education in the area of ​​\u200b\u200bthe problems being solved increases.

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

Rules for entering numbers

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

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

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

The denominator cannot be negative.

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

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

Enter numbers a n , d, n

Find a 1

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A little theory.

Number sequence

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

In a savings bank, using the depositor’s personal account number, you can easily find this account and see what deposit is on it. Let account No. 1 contain a deposit of a1 rubles, account No. 2 contain a deposit of a2 rubles, etc. It turns out number sequence
a 1 , a 2 , a 3 , ..., a N
where N is the number of all accounts. Here, each natural number n from 1 to N is associated with a number a n.

Also studied in mathematics infinite number sequences:
a 1 , a 2 , a 3 , ..., a n , ... .
The number a 1 is called first term of the sequence, number a 2 - second term of the sequence, number a 3 - third term of the sequence etc.
The number a n is called nth (nth) member of the sequence, and the natural number n is its number.

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

Arithmetic progression

The length of the year is approximately 365 days. A more accurate value is \(365\frac(1)(4)\) days, so every four years an error of one day accumulates.

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

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

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

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

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

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

Thus, each term of an arithmetic progression, starting from the second, is equal to the arithmetic mean of its two adjacent terms. This explains the name "arithmetic" progression.

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

Sum of the first n terms of an arithmetic progression

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

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

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

Books (textbooks) Abstracts of the Unified State Examination and the Unified State Examination tests online Games, puzzles Plotting graphs of functions Spelling dictionary of the Russian language Dictionary of youth slang Catalog of Russian schools Catalog of secondary educational institutions of Russia Catalog of Russian universities List of tasks

When studying algebra in a secondary school (grade 9) one of important topics is the study number sequences, which include progressions - geometric and arithmetic. In this article we will look at an arithmetic progression and examples with solutions.

What is an arithmetic progression?

To understand this, it is necessary to define the progression in question, as well as provide the basic formulas that will be used later in solving problems.

Arithmetic or is a set of ordered rational numbers, each member of which differs from the previous one by some constant value. This value is called the difference. That is, knowing any member of an ordered series of numbers and the difference, you can restore the entire arithmetic progression.

Let's give an example. The following sequence of numbers will be an arithmetic progression: 4, 8, 12, 16, ..., since the difference in this case is 4 (8 - 4 = 12 - 8 = 16 - 12). But the set of numbers 3, 5, 8, 12, 17 can no longer be attributed to the type of progression under consideration, since the difference for it is not a constant value (5 - 3 ≠ 8 - 5 ≠ 12 - 8 ≠ 17 - 12).

Important Formulas

Let us now present the basic formulas that will be needed to solve problems using arithmetic progression. Let us denote by the symbol a n the nth member of the sequence, where n is an integer. We denote the difference by the Latin letter d. Then the following expressions are valid:

1. To determine the value of the nth term, the following formula is suitable: a n = (n-1)*d+a 1 .
2. To determine the sum of the first n terms: S n = (a n +a 1)*n/2.

To understand any examples of arithmetic progression with solutions in 9th grade, it is enough to remember these two formulas, since any problems of the type under consideration are based on their use. You should also remember that the progression difference is determined by the formula: d = a n - a n-1.

Example #1: finding an unknown term

Let's give a simple example of an arithmetic progression and the formulas that need to be used to solve it.

Let the sequence 10, 8, 6, 4, ... be given, you need to find five terms in it.

From the conditions of the problem it already follows that the first 4 terms are known. The fifth can be defined in two ways:

1. Let's first calculate the difference. We have: d = 8 - 10 = -2. Similarly, you could take any two other members standing next to each other. For example, d = 4 - 6 = -2. Since it is known that d = a n - a n-1, then d = a 5 - a 4, from which we get: a 5 = a 4 + d. We substitute the known values: a 5 = 4 + (-2) = 2.
2. The second method also requires knowledge of the difference of the progression in question, so you first need to determine it as shown above (d = -2). Knowing that the first term a 1 = 10, we use the formula for the n number of the sequence. We have: a n = (n - 1) * d + a 1 = (n - 1) * (-2) + 10 = 12 - 2*n. Substituting n = 5 into the last expression, we get: a 5 = 12-2 * 5 = 2.

As you can see, both solutions led to the same result. Note that in this example the progression difference d is a negative value. Such sequences are called decreasing, since each next term is less than the previous one.

Example #2: progression difference

Now let's complicate the task a little, give an example of how to find the difference of an arithmetic progression.

It is known that in some algebraic progression the 1st term is equal to 6, and the 7th term is equal to 18. It is necessary to find the difference and restore this sequence to the 7th term.

Let's use the formula to determine the unknown term: a n = (n - 1) * d + a 1 . Let's substitute the known data from the condition into it, that is, the numbers a 1 and a 7, we have: 18 = 6 + 6 * d. From this expression you can easily calculate the difference: d = (18 - 6) /6 = 2. Thus, we have answered the first part of the problem.

To restore the sequence to the 7th term, you should use the definition algebraic progression, that is, a 2 = a 1 + d, a 3 = a 2 + d and so on. As a result, we restore the entire sequence: a 1 = 6, a 2 = 6 + 2=8, a 3 = 8 + 2 = 10, a 4 = 10 + 2 = 12, a 5 = 12 + 2 = 14, a 6 = 14 + 2 = 16, a 7 = 18.

Example No. 3: drawing up a progression

Let's complicate it further stronger condition tasks. Now we need to answer the question of how to find an arithmetic progression. The following example can be given: two numbers are given, for example - 4 and 5. It is necessary to create an algebraic progression so that three more terms are placed between these.

Before you start solving this problem, you need to understand what place the given numbers will occupy in the future progression. Since there will be three more terms between them, then a 1 = -4 and a 5 = 5. Having established this, we move on to the problem, which is similar to the previous one. Again, for the nth term we use the formula, we get: a 5 = a 1 + 4 * d. From: d = (a 5 - a 1)/4 = (5 - (-4)) / 4 = 2.25. What we got here is not an integer value of the difference, but it is a rational number, so the formulas for the algebraic progression remain the same.

Now let's add the found difference to a 1 and restore the missing terms of the progression. We get: a 1 = - 4, a 2 = - 4 + 2.25 = - 1.75, a 3 = -1.75 + 2.25 = 0.5, a 4 = 0.5 + 2.25 = 2.75, a 5 = 2.75 + 2.25 = 5, which coincided with the conditions of the problem.

Example No. 4: first term of progression

Let's continue to give examples of arithmetic progression with solutions. In all previous problems, the first number of the algebraic progression was known. Now let's consider a problem of a different type: let two numbers be given, where a 15 = 50 and a 43 = 37. It is necessary to find which number this sequence begins with.

The formulas used so far assume knowledge of a 1 and d. In the problem statement, nothing is known about these numbers. Nevertheless, we will write down expressions for each term about which information is available: a 15 = a 1 + 14 * d and a 43 = a 1 + 42 * d. We received two equations in which there are 2 unknown quantities (a 1 and d). This means that the problem is reduced to solving a system of linear equations.

The easiest way to solve this system is to express a 1 in each equation and then compare the resulting expressions. First equation: a 1 = a 15 - 14 * d = 50 - 14 * d; second equation: a 1 = a 43 - 42 * d = 37 - 42 * d. Equating these expressions, we get: 50 - 14 * d = 37 - 42 * d, whence the difference d = (37 - 50) / (42 - 14) = - 0.464 (only 3 decimal places are given).

Knowing d, you can use any of the 2 expressions above for a 1. For example, first: a 1 = 50 - 14 * d = 50 - 14 * (- 0.464) = 56.496.

If you have doubts about the result obtained, you can check it, for example, determine the 43rd term of the progression, which is specified in the condition. We get: a 43 = a 1 + 42 * d = 56.496 + 42 * (- 0.464) = 37.008. The small error is due to the fact that rounding to thousandths was used in the calculations.

Example No. 5: amount

Now let's look at several examples with solutions for the sum of an arithmetic progression.

Let a numerical progression of the following form be given: 1, 2, 3, 4, ...,. How to calculate the sum of 100 of these numbers?

Thanks to the development of computer technology, it is possible to solve this problem, that is, add all the numbers sequentially, which the computer will do as soon as a person presses the Enter key. However, the problem can be solved mentally if you pay attention that the presented series of numbers is an algebraic progression, and its difference is equal to 1. Applying the formula for the sum, we get: S n = n * (a 1 + a n) / 2 = 100 * (1 + 100) / 2 = 5050.

It is interesting to note that this problem is called “Gaussian” because at the beginning of the 18th century the famous German, still only 10 years old, was able to solve it in his head in a few seconds. The boy did not know the formula for the sum of an algebraic progression, but he noticed that if you add the numbers at the ends of the sequence in pairs, you always get the same result, that is, 1 + 100 = 2 + 99 = 3 + 98 = ..., and since these sums will be exactly 50 (100 / 2), then to get the correct answer it is enough to multiply 50 by 101.

Example No. 6: sum of terms from n to m

Another typical example of the sum of an arithmetic progression is the following: given a series of numbers: 3, 7, 11, 15, ..., you need to find what the sum of its terms from 8 to 14 will be equal to.

The problem is solved in two ways. The first of them involves finding unknown terms from 8 to 14, and then summing them sequentially. Since there are few terms, this method is not quite labor-intensive. Nevertheless, it is proposed to solve this problem using a second method, which is more universal.

The idea is to obtain a formula for the sum of the algebraic progression between terms m and n, where n > m are integers. For both cases, we write two expressions for the sum:

1. S m = m * (a m + a 1) / 2.
2. S n = n * (a n + a 1) / 2.

Since n > m, it is obvious that the 2nd sum includes the first. The last conclusion means that if we take the difference between these sums and add the term a m to it (in the case of taking the difference, it is subtracted from the sum S n), we will obtain the necessary answer to the problem. We have: S mn = S n - S m + a m =n * (a 1 + a n) / 2 - m *(a 1 + a m)/2 + a m = a 1 * (n - m) / 2 + a n * n/2 + a m * (1- m/2). It is necessary to substitute formulas for a n and a m into this expression. Then we get: S mn = a 1 * (n - m) / 2 + n * (a 1 + (n - 1) * d) / 2 + (a 1 + (m - 1) * d) * (1 - m / 2) = a 1 * (n - m + 1) + d * n * (n - 1) / 2 + d *(3 * m - m 2 - 2) / 2.

The resulting formula is somewhat cumbersome, however, the sum S mn depends only on n, m, a 1 and d. In our case, a 1 = 3, d = 4, n = 14, m = 8. Substituting these numbers, we get: S mn = 301.

As can be seen from the above solutions, all problems are based on knowledge of the expression for the nth term and the formula for the sum of the set of first terms. Before starting to solve any of these problems, it is recommended that you carefully read the condition, clearly understand what you need to find, and only then proceed with the solution.

Another tip is to strive for simplicity, that is, if you can answer a question without using complex mathematical calculations, then you need to do just that, since in this case the likelihood of making a mistake is less. For example, in the example of an arithmetic progression with solution No. 6, one could stop at the formula S mn = n * (a 1 + a n) / 2 - m * (a 1 + a m) / 2 + a m, and divide the overall problem into separate subtasks (in this case, first find the terms a n and a m).

If you have doubts about the result obtained, it is recommended to check it, as was done in some of the examples given. We found out how to find an arithmetic progression. If you figure it out, it's not that difficult.

Lesson type: learning new material.

Lesson objectives:

• expanding and deepening students’ understanding of problems solved using arithmetic progression; organizing students' search activities when deriving the formula for the sum of the first n terms of an arithmetic progression;
• developing the ability to independently acquire new knowledge and use already acquired knowledge to achieve a given task;
• developing the desire and need to generalize the facts obtained, developing independence.

Tasks:

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

Equipment:

• cards with tasks for working in groups and pairs;
• evaluation paper;
• presentation“Arithmetic progression.”

I. Updating of basic knowledge.

1. Independent work in pairs.

1st option:

Define arithmetic progression. Write down a recurrence formula that defines an arithmetic progression. Please provide 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 the arithmetic progression ( a n}: 2, 5, 8 …
At this time, two students back side boards are preparing answers to these same questions.
Students evaluate their partner's work by checking them on the board. (Sheets with answers are handed in.)

2. Game moment.

Exercise 1.

Teacher. I thought of some arithmetic progression. Ask me only two questions so that after the answers you can quickly name the 7th term 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 equal to. You can ask questions: what is the 6th term of the progression equal to and what is the 8th term of the progression equal to?

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 board. Students call out the number, and the teacher instantly calls out the number itself. Explain how I can do this?

The teacher remembers the formula for the nth term a n = 3n – 2 and, substituting the specified values ​​n, finds the corresponding values a n.

II. Setting a learning task.

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

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

• How is this problem related to the topic arithmetic progression? (Each next person receives 1/8 of the measure more, which means the difference is d=1/8, 10 people, which means n=10.)
• What do you think the number 10 measures means? (Sum of all terms of the progression.)
• What else do you need to know to make it easy and simple to divide the barley according to the conditions of the problem? (First term of 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 we deduce the formula, let's look at how the ancient Egyptians solved the problem.

And they solved it as follows:

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

We get the sequence:

III. Solving the problem.

1. Work in groups

Group I: Find the sum of 20 consecutive natural numbers: S 20 =(20+1)∙10 =210.

In general

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

S 100 = (1+100)∙50 = 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 problems considered 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 =a 1 + a 2 + a 3 + a 4 +…+ a n-3 + a n-2 + a n-1 + a n.

Let's find this sum using similar reasoning:

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

IV. Primary understanding and application of the obtained formulas when solving problems.

1. Checking the solution to an ancient problem using the formula.

2. Application of the formula in solving various problems.

3. Exercises to develop the ability to apply formulas when solving problems.

A) No. 613

Given: ( a n) – arithmetic progression;

(a n): 1, 2, 3, …, 1500

Find: S 1500

Solution: , a 1 = 1, and 1500 = 1500,

B) Given: ( a n) – arithmetic progression;
(a n): 1, 2, 3, …
S n = 210

Find: n
Solution:

V. Independent work with mutual verification.

Denis started working 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 total in a year?

Given: ( a n) – arithmetic progression;
a 1 = 200, d=30, n=12
Find: S 12
Solution:

Answer: Denis received 4380 rubles for the year.

VI. Homework instruction.

1. Section 4.3 – learn the derivation of the formula.
2. №№ 585, 623 .
3. Create a problem that can 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...
• Formulas learned...
• I believe 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. Tutorial for educational institutions. Ed. G.V. Dorofeeva. M.: “Enlightenment”, 2009.