Arithmetic progression formula how to find an. Arithmetic progression

IV Yakovlev | Materials on mathematics | MathUs.ru

An arithmetic progression is a special kind of sequence. Therefore, before defining an arithmetic (and then geometric) progression, we need to briefly discuss the important concept of a number sequence.

Subsequence

Imagine a device on the screen of which some numbers are displayed one after another. Let's say 2; 7; 13; one; 6; 0; 3; : : : Such a set of numbers is just an example of a sequence.

Definition. A numerical sequence is a set of numbers in which each number can be assigned a unique number (that is, put in correspondence with a single natural number)1. The number with number n is called nth member sequences.

So, in the above example, the first number has the number 2, which is the first member of the sequence, which can be denoted by a1 ; the number five has the number 6 which is the fifth member of the sequence, which can be denoted a5 . Generally, nth member sequences are denoted by an (or bn , cn , etc.).

A very convenient situation is when the nth member of the sequence can be specified by some formula. For example, the formula an = 2n 3 specifies the sequence: 1; one; 3; 5; 7; : : : The formula an = (1)n defines the sequence: 1; one; one; one; : : :

Not every set of numbers is a sequence. So, a segment is not a sequence; it contains ¾too many¿ numbers to be renumbered. The set R of all real numbers is also not a sequence. These facts are proved in the course of mathematical analysis.

Arithmetic progression: basic definitions

Now we are ready to define an arithmetic progression.

Definition. An arithmetic progression is a sequence in which each term (starting from the second) is equal to the sum the previous term and some fixed number (called the difference of an arithmetic progression).

For example, sequence 2; 5; eight; eleven; : : : is an arithmetic progression with first term 2 and difference 3. Sequence 7; 2; 3; eight; : : : is an arithmetic progression with first term 7 and difference 5. Sequence 3; 3; 3; : : : is an arithmetic progression with zero difference.

Equivalent definition: A sequence an is called an arithmetic progression if the difference an+1 an is a constant value (not dependent on n).

An arithmetic progression is said to be increasing if its difference is positive, and decreasing if its difference is negative.

1 And here is a more concise definition: a sequence is a function defined on a set natural numbers. For example, the sequence of real numbers is the function f: N! R.

By default, sequences are considered infinite, that is, containing an infinite number of numbers. But no one bothers to consider finite sequences as well; in fact, any finite set of numbers can be called a finite sequence. For example, the final sequence 1; 2; 3; four; 5 consists of five numbers.

Formula of the nth member of an arithmetic progression

It is easy to understand that an arithmetic progression is completely determined by two numbers: the first term and the difference. Therefore, the question arises: how, knowing the first term and the difference, find an arbitrary term of an arithmetic progression?

It is not difficult to obtain the desired formula for the nth term of an arithmetic progression. Let an

arithmetic progression with difference d. We have:
an+1 = an + d (n = 1; 2; : ::):
In particular, we write:
a2 = a1 + d;
a3 = a2 + d = (a1 + d) + d = a1 + 2d;
a4 = a3 + d = (a1 + 2d) + d = a1 + 3d;
and now it becomes clear that the formula for an is:
an = a1 + (n 1)d:

Task 1. In arithmetic progression 2; 5; eight; eleven; : : : find the formula of the nth term and calculate the hundredth term.

Solution. According to formula (1) we have:

an = 2 + 3(n 1) = 3n 1:

a100 = 3 100 1 = 299:

Property and sign of arithmetic progression

property of an arithmetic progression. In arithmetic progression an for any

In other words, each member of the arithmetic progression (starting from the second) is the arithmetic mean of the neighboring members.

	Proof. We have:
	a n 1 + a n+1		(an d) + (an + d)

which is what was required.

More generally, the arithmetic progression an satisfies the equality

a n = a n k + a n+k

for any n > 2 and any natural k< n. Попробуйте самостоятельно доказать эту формулу тем же самым приёмом, что и формулу (2 ).

It turns out that formula (2) is not only a necessary but also a sufficient condition for a sequence to be an arithmetic progression.

Sign of an arithmetic progression. If equality (2) holds for all n > 2, then the sequence an is an arithmetic progression.

Proof. Let's rewrite the formula (2) as follows:

a n a n 1 = a n+1 a n:

This shows that the difference an+1 an does not depend on n, and this just means that the sequence an is an arithmetic progression.

The property and sign of an arithmetic progression can be formulated as one statement; for convenience, we will do this for three numbers (this is the situation that often occurs in problems).

Characterization of an arithmetic progression. Three numbers a, b, c form an arithmetic progression if and only if 2b = a + c.

Problem 2. (Moscow State University, Faculty of Economics, 2007) Three numbers 8x, 3 x2 and 4 in the specified order form a decreasing arithmetic progression. Find x and write the difference of this progression.

Solution. By the property of an arithmetic progression, we have:

2(3 x2 ) = 8x 4 , 2x2 + 8x 10 = 0 , x2 + 4x 5 = 0 , x = 1; x=5:

If x = 1, then a decreasing progression of 8, 2, 4 is obtained with a difference of 6. If x = 5, then an increasing progression of 40, 22, 4 is obtained; this case does not work.

Answer: x = 1, the difference is 6.

The sum of the first n terms of an arithmetic progression

The legend says that once the teacher told the children to find the sum of numbers from 1 to 100 and sat down to read the newspaper quietly. However, within a few minutes, one boy said that he had solved the problem. It was 9-year-old Carl Friedrich Gauss, later one of the greatest mathematicians in history.

Little Gauss's idea was this. Let

S = 1 + 2 + 3 + : : : + 98 + 99 + 100:

Let's write this sum in reverse order:

S = 100 + 99 + 98 + : : : + 3 + 2 + 1;

and add these two formulas:

2S = (1 + 100) + (2 + 99) + (3 + 98) + : : : + (98 + 3) + (99 + 2) + (100 + 1):

Each term in brackets is equal to 101, and there are 100 such terms in total. Therefore

2S = 101 100 = 10100;

We use this idea to derive the sum formula

S = a1 + a2 + : : : + an + a n n: (3)

A useful modification of formula (3) is obtained by substituting the formula for the nth term an = a1 + (n 1)d into it:

		2a1 + (n 1)d

Task 3. Find the sum of all positive three-digit numbers divisible by 13.

Solution. Three-digit numbers that are multiples of 13 form an arithmetic progression with the first term 104 and the difference 13; The nth term of this progression is:

an = 104 + 13(n 1) = 91 + 13n:

Let's find out how many members our progression contains. To do this, we solve the inequality:

an 6999; 91 + 13n 6999;

n 6 908 13 = 6911 13; n 6 69:

So there are 69 members in our progression. According to the formula (4) we find the required amount:

S = 2 104 + 68 13 69 = 37674: 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Definition. An arithmetic progression is called:

increasing if each next element is greater than the previous one;

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:

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

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

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

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

Members of the progression and the recurrent formula

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

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

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

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

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

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

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

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

However, I suggest you practice a little.

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

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

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

Answer: (8; 3; -2)

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

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

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

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

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

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

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

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

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

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

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

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

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

Ready! Problem solved.

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

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

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

Simple but very useful property, which you definitely need to know - with its help you can significantly speed up the solution of many problems in progressions. Here bright to that example:

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

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

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

Answer: 20.4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Arithmetic mean and equal indents

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

Arithmetic progression members on the number line

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

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

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

However, these equalities can be rewritten differently:

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

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

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

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

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

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

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

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

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

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

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

It turned out classic quadratic equation. Its roots: $x=2$ and $x=-3$ are the answers.

Answer: -3; 2.

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

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

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

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

Answer: 1; 6.

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

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

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

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

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

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

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

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

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

Grouping and sum of elements

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

6 elements marked on the number line

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

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

Now note that the following sums are equal:

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

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

Same indents give equal sums

Understanding this fact will allow us to solve problems fundamentally more high level complexity than those discussed above. For example, these:

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

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 difference of the progression $d$. Actually, the whole solution will be built around the difference, since the product $((a)_(2))\cdot ((a)_(12))$ can be rewritten as follows:

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

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

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

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

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

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

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

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

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

Answer: -36

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

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

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

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

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

Arguing similarly, we find the remaining number:

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

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

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

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

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

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

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

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

But then the above expression can be rewritten like this:

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

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

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

It remains only to find the remaining members:

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

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

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

Text tasks with progressions

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

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

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

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

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

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

Therefore, 202 parts will be manufactured in November.

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

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

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

Numeric sequence
For example, for our sequence:

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

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

In our case:

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

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

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

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

a)
b)
c)
d)

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

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

1. Method

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

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

2. Way

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

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

In other words:

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

Calculated? Compare your entries with the answer:

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

Arithmetic progression equation.

Arithmetic progressions are either increasing or decreasing.

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

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

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

Since then:

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

Let's compare the results:

Arithmetic progression property

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

Let, a, then:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Method 1.

Method 2.

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

Workout

Tasks:

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

Answers:

Let us define the parameters of the arithmetic progression. In this case
(weeks = days).
Answer: In two weeks, Masha should squat once a day.
First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in - half, however, check this fact using the formula for finding the -th member of an arithmetic progression:
The numbers do contain odd numbers.
We substitute the available data into the formula:
Answer: The sum of all odd numbers contained in is equal to.
Recall the problem about the pyramids. For our case, a , since each top layer is reduced by one log, there are only a bunch of layers, that is.
Substitute the data in the formula:
Answer: There are logs in the masonry.

Summing up

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

, where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Numeric sequence

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

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

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

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

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

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

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

sets the sequence:

And the formula is the following sequence:

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

nth term formula

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

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

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

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

Much more comfortable now, right? We check:

Decide for yourself:

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

Solution:

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

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

So the formula is:

Then the hundredth term is:

What is the sum of all natural numbers from to?

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

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

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

Solution:

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

The formula for the th term for this progression is:

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

Very easy: .

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

Answer: .

Now decide for yourself:

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

Answers:

The most important thing here is to recognize the arithmetic progression and determine its parameters. In this case, (weeks = days). You need to determine the sum of the first terms of this progression:
.
Answer:
Here it is given:, it is necessary to find.
Obviously, you need to use the same sum formula as in the previous problem:
.
Substitute the values:
The root obviously doesn't fit, so the answer.
Let's calculate the distance traveled over the last day using the formula of the -th member:
(km).
Answer:
Given: . Find: .
It doesn't get easier:
(rub).
Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN

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

Arithmetic progression is increasing () and decreasing ().

For example:

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

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

Property of members of an arithmetic progression

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

The sum of the members of an arithmetic progression

There are two ways to find the sum:

Where is the number of values.

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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.

What is the essence of the formula?

This formula allows you to find any BY HIS NUMBER" n" .

Of course, you need to know the first term a 1 and progression difference d, well, without these parameters, you can’t write down a specific progression.

It is not enough to memorize (or cheat) this formula. It is necessary to assimilate its essence and apply the formula in various problems. Yes, and do not forget at the right time, yes ...) How not forget- I do not know. But how to remember If needed, I'll give you a hint. For those who master the lesson to the end.)

So, let's deal with the formula of the n-th member of an arithmetic progression.

What is a formula in general - we imagine.) What is an arithmetic progression, a member number, a progression difference - is clearly stated in the previous lesson. Take a look if you haven't read it. Everything is simple there. It remains to figure out what nth member.

progression in general view can be written as a series of numbers:

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

a 1- denotes the first term of an arithmetic progression, a 3- third member a 4- fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - from a 120.

How to define in general any member of an arithmetic progression, s any number? Very simple! Like this:

a n

That's what it is n-th member of an arithmetic progression. Under the letter n all the numbers of members are hidden at once: 1, 2, 3, 4, and so on.

And what does such a record give us? Just think, instead of a number, they wrote down a letter ...

This notation gives us a powerful tool for working with arithmetic progressions. Using the notation a n, we can quickly find any member any arithmetic progression. And a bunch of tasks to solve in progression. You will see further.

In the formula of the nth member of an arithmetic progression:

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

a 1- the first member of the arithmetic progression;

n- member number.

The formula links the key parameters of any progression: a n ; a 1 ; d and n. Around these parameters, all the puzzles revolve in progression.

The nth term formula can also be used to write a specific progression. For example, in the problem it can be said that the progression is given by the condition:

a n = 5 + (n-1) 2.

Such a problem can even confuse ... There is no series, no difference ... But, comparing the condition with the formula, it is easy to figure out that in this progression a 1 \u003d 5, and d \u003d 2.

And it can be even angrier!) If we take the same condition: a n = 5 + (n-1) 2, yes, open the brackets and give similar ones? We get a new formula:

an = 3 + 2n.

it Only not general, but for a specific progression. This is where the pitfall lies. Some people think that the first term is a three. Although in reality the first member is a five ... A little lower we will work with such a modified formula.

In tasks for progression, there is another notation - a n+1. This is, you guessed it, the "n plus the first" term of the progression. Its meaning is simple and harmless.) This is a member of the progression, the number of which is greater than the number n by one. For example, if in some problem we take for a n fifth term, then a n+1 will be the sixth member. Etc.

Most often the designation a n+1 occurs in recursive formulas. Do not be afraid of this terrible word!) This is just a way of expressing a term of an arithmetic progression through the previous one. Suppose we are given an arithmetic progression in this form, using the recurrent formula:

a n+1 = a n +3

a 2 = a 1 + 3 = 5+3 = 8

a 3 = a 2 + 3 = 8+3 = 11

The fourth - through the third, the fifth - through the fourth, and so on. And how to count immediately, say the twentieth term, a 20? But no way!) While the 19th term is not known, the 20th cannot be counted. This is the fundamental difference between the recursive formula and the formula of the nth term. Recursive works only through previous term, and the formula of the nth term - through the first and allows straightaway find any member by its number. Not counting the whole series of numbers in order.

In an arithmetic progression, a recursive formula can easily be turned into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in the usual form, and work with it. In the GIA, such tasks are often found.

Application of the formula of the n-th member of an arithmetic progression.

First, let's look at the direct application of the formula. At the end of the previous lesson there was a problem:

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

This problem can be solved without any formulas, simply based on the meaning of the arithmetic progression. Add, yes add ... An hour or two.)

And according to the formula, the solution will take less than a minute. You can time it.) We decide.

The conditions provide all the data for using the formula: a 1 \u003d 3, d \u003d 1/6. It remains to be seen what n. No problem! We need to find a 121. Here we write:

Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in the member of the arithmetic progression number one hundred twenty one. This will be our n. It is this meaning n= 121 we will substitute further into the formula, in brackets. Substitute all the numbers in the formula and calculate:

a 121 = 3 + (121-1) 1/6 = 3+20 = 23

That's all there is to it. Just as quickly one could find the five hundred and tenth member, and the thousand and third, any. We put instead n the desired number in the index of the letter " a" and in brackets, and we consider.

Let me remind you the essence: this formula allows you to find any term of an arithmetic progression BY HIS NUMBER" n" .

Let's solve the problem smarter. Let's say we have the following problem:

Find the first term of the arithmetic progression (a n) if a 17 =-2; d=-0.5.

If you have any difficulties, I will suggest the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Hand write, right in your notebook:

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

And now, looking at the letters of the formula, we understand what data we have and what is missing? Available d=-0.5, there is a seventeenth member ... Everything? If you think that's all, then you can't solve the problem, yes ...

We also have a number n! In the condition a 17 =-2 hidden two options. This is both the value of the seventeenth member (-2) and its number (17). Those. n=17. This "little thing" often slips past the head, and without it, (without the "little thing", not the head!) The problem cannot be solved. Although ... and without a head too.)

Now we can just stupidly substitute our data into the formula:

a 17 \u003d a 1 + (17-1) (-0.5)

Oh yes, a 17 we know it's -2. Okay, let's put it in:

-2 \u003d a 1 + (17-1) (-0.5)

That, in essence, is all. It remains to express the first term of the arithmetic progression from the formula, and calculate. You get the answer: a 1 = 6.

Such a technique - writing a formula and simply substituting known data - helps a lot in simple tasks. Well, you must, of course, be able to express a variable from a formula, but what to do!? Without this skill, mathematics can not be studied at all ...

Another popular problem:

Find the difference of the arithmetic progression (a n) if a 1 =2; a 15 =12.

What are we doing? You will be surprised, we write the formula!)

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

Consider what we know: a 1 =2; a 15 =12; and (special highlight!) n=15. Feel free to substitute in the formula:

12=2 + (15-1)d

Let's do the arithmetic.)

12=2 + 14d

d=10/14 = 5/7

This is the correct answer.

So, tasks a n , a 1 and d decided. It remains to learn how to find the number:

The number 99 is a member of an arithmetic progression (a n), where a 1 =12; d=3. Find the number of this member.

We substitute the known quantities into the formula of the nth term:

a n = 12 + (n-1) 3

At first glance, there are two unknown quantities here: a n and n. But a n is some member of the progression with the number n... And this member of the progression we know! It's 99. We don't know his number. n, so this number also needs to be found. Substitute the progression term 99 into the formula:

99 = 12 + (n-1) 3

We express from the formula n, we think. We get the answer: n=30.

And now a problem on the same topic, but more creative):

Determine if the number 117 will be a member of an arithmetic progression (a n):

-3,6; -2,4; -1,2 ...

Let's write the formula again. What, there are no options? Hm... Why do we need eyes?) Do we see the first member of the progression? We see. This is -3.6. You can safely write: a 1 \u003d -3.6. Difference d can be determined from the series? It's easy if you know what the difference of an arithmetic progression is:

d = -2.4 - (-3.6) = 1.2

Yes, we did the simplest thing. It remains to deal with an unknown number n and an incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know that ... How to be!? Well, how to be, how to be... Turn on Creative skills!)

We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes-yes!)) and substitute our numbers:

117 = -3.6 + (n-1) 1.2

Again we express from the formulan, we count and get:

Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion do we draw? Yes! Number 117 is not member of our progression. It is somewhere between the 101st and 102nd members. If the number turned out to be natural, i.e. positive integer, then the number would be a member of the progression with the found number. And in our case, the answer to the problem will be: no.

Task based real version GIA:

The arithmetic progression is given by the condition:

a n \u003d -4 + 6.8n

Find the first and tenth terms of the progression.

Here the progression is set in an unusual way. Some kind of formula ... It happens.) However, this formula (as I wrote above) - also the formula of the n-th member of an arithmetic progression! She also allows find any member of the progression by its number.

We are looking for the first member. The one who thinks. that the first term is minus four, is fatally mistaken!) Because the formula in the problem is modified. The first term of an arithmetic progression in it hidden. Nothing, we'll find it now.)

Just as in the previous tasks, we substitute n=1 into this formula:

a 1 \u003d -4 + 6.8 1 \u003d 2.8

Here! The first term is 2.8, not -4!

Similarly, we are looking for the tenth term:

a 10 \u003d -4 + 6.8 10 \u003d 64

That's all there is to it.

And now, for those who have read up to these lines, the promised bonus.)

Suppose, in a difficult combat situation, the GIA or the Unified State Examination, you forgot useful formula nth member of an arithmetic progression. Something comes to mind, but somehow uncertainly ... Whether n there, or n+1, or n-1... How to be!?

Calm! This formula is easy to derive. Not very strict, but definitely enough for confidence and the right decision!) For the conclusion, it is enough to remember the elementary meaning of the arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.

We draw numerical axis and mark the first one on it. second, third, etc. members. And note the difference d between members. Like this:

We look at the picture and think: what is the second term equal to? Second one d:

a 2 =a 1 + 1 d

What is the third term? Third term equals first term plus two d.

a 3 =a 1 + 2 d

Do you get it? I don't put some words in bold for nothing. Okay, one more step.)

What is the fourth term? Fourth term equals first term plus three d.

a 4 =a 1 + 3 d

It's time to realize that the number of gaps, i.e. d, always one less than the number of the member you are looking for n. That is, up to the number n, number of gaps will be n-1. So, the formula will be (no options!):

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

In general, visual pictures are very helpful in solving many problems in mathematics. Don't neglect the pictures. But if it's difficult to draw a picture, then ... only a formula!) In addition, the formula of the nth term allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. You can't put a picture in an equation...

Tasks for independent decision.

For warm-up:

1. In arithmetic progression (a n) a 2 =3; a 5 \u003d 5.1. Find a 3 .

Hint: according to the picture, the problem is solved in 20 seconds ... According to the formula, it turns out more difficult. But for mastering the formula, it is more useful.) In Section 555, this problem is solved both by the picture and by the formula. Feel the difference!)

And this is no longer a warm-up.)

2. In arithmetic progression (a n) a 85 \u003d 19.1; a 236 =49, 3. Find a 3 .

What, reluctance to draw a picture?) Still! It's better in the formula, yes ...

3. Arithmetic progression is given by the condition:a 1 \u003d -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.

In this task, the progression is given in a recurrent way. But counting up to the one hundred and twenty-fifth term... Not everyone can do such a feat.) But the formula of the nth term is within the power of everyone!

4. Given an arithmetic progression (a n):

-148; -143,8; -139,6; -135,4, .....

Find the number of the smallest positive term of the progression.

5. According to the condition of task 4, find the sum of the smallest positive and largest negative members of the progression.

6. The product of the fifth and twelfth terms of an increasing arithmetic progression is -2.5, and the sum of the third and eleventh terms is zero. Find a 14 .

Not the easiest task, yes ...) Here the method "on the fingers" will not work. You have to write formulas and solve equations.

Answers (in disarray):

3,7; 3,5; 2,2; 37; 2,7; 56,5

Happened? It's nice!)

Not everything works out? It happens. By the way, in the last task there is one subtle point. Attentiveness when reading the problem will be required. And logic.

The solution to all these problems is discussed in detail in Section 555. And the fantasy element for the fourth, and the subtle moment for the sixth, and general approaches for solving any problems for the formula of the nth term - everything is painted. I recommend.

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