The sum of arithmetic and geometric. Arithmetic progression – number sequence

17.10.2019

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

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

The first element of the sequence;

The fifth element of the sequence;

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

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

The sequence can be set in three ways:

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

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

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

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

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

For example, if , then

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

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

If, for example, , That

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

3 . The sequence can be specified using a formula that expresses the dependence of the value of the sequence member number n on the values of the previous members.

In this case, it is not enough for us to know only the number of the sequence member to find its value. We need to specify the first member or first few members of the sequence. ,

For example, consider the sequence We can find the values of sequence members in sequence

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

- come back.

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

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

. The difference of an arithmetic progression can be positive, negative, or equal to zero.">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} If title="d>0.

increasing

For example, 2; 5; 8; eleven;... If , then each term of an arithmetic progression is less than the previous one, and the progression is.

decreasing

For example, 2; -1; -4; -7;... If , then all terms of the progression are equal to the same number, and the progression is.

stationary

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

The main property of an arithmetic progression:

Let's look at the drawing.

We see that

, and at the same time

Adding these two equalities, we get:

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

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

We see that

Moreover, since

, That

, and therefore">, равен среднему арифметическому двух равноотстоящих. !}

Each term of an arithmetic progression, starting with title="k>l

Formula of the th term.

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

and finally We got

formula of the nth term. IMPORTANT!

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

The sum of n terms of an arithmetic progression.

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

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

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

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

We get:

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

Let's consider solving arithmetic progression problems.

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

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

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

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

a) Find 31 terms of the progression.

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

A) We see that ;

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

In general

In our case , That's why

If for every natural number n match a real number a n , then they say that it is given number sequence :

a 1 , a 2 , a 3 , . . . , a n , . . . .

So, the number sequence is a function of the natural argument.

Number a 1 called first term of the sequence , number a 2 — second term of the sequence , number a 3 — third and so on. Number a n called nth member of the sequence , and a natural number n — his number .

From two adjacent members a n And a n +1 sequence member a n +1 called subsequent (towards a n ), A a n — previous (towards a n +1 ).

To define a sequence, you need to specify a method that allows you to find a member of the sequence with any number.

Often the sequence is specified using nth term formulas , that is, a formula that allows you to determine a member of a sequence by its number.

For example,

a sequence of positive odd numbers can be given by the formula

a n= 2n- 1,

and the sequence of alternating 1 And -1 - formula

b n = (-1)n +1 . ◄

The sequence can be determined recurrent formula, that is, a formula that expresses any member of the sequence, starting with some, through the previous (one or more) members.

For example,

If a 1 = 1 , A a n +1 = a n + 5

a 1 = 1,

a 2 = a 1 + 5 = 1 + 5 = 6,

a 3 = a 2 + 5 = 6 + 5 = 11,

a 4 = a 3 + 5 = 11 + 5 = 16,

a 5 = a 4 + 5 = 16 + 5 = 21.

If a 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven terms of the numerical sequence are established as follows:

a 1 = 1,

a 2 = 1,

a 3 = a 1 + a 2 = 1 + 1 = 2,

a 4 = a 2 + a 3 = 1 + 2 = 3,

a 5 = a 3 + a 4 = 2 + 3 = 5,

a 6 = a 4 + a 5 = 3 + 5 = 8,

a 7 = a 5 + a 6 = 5 + 8 = 13. ◄

Sequences can be final And endless .

The sequence is called ultimate , if it has a finite number of members. The sequence is called endless , if it has infinitely many members.

For example,

sequence of two-digit natural numbers:

10, 11, 12, 13, . . . , 98, 99

final.

Sequence of prime numbers:

2, 3, 5, 7, 11, 13, . . .

endless. ◄

The sequence is called increasing , if each of its members, starting from the second, is greater than the previous one.

The sequence is called decreasing , if each of its members, starting from the second, is less than the previous one.

For example,

2, 4, 6, 8, . . . , 2n, . . . — increasing sequence;

1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . — decreasing sequence. ◄

A sequence whose elements do not decrease as the number increases, or, conversely, do not increase, is called monotonous sequence .

Monotonic sequences, in particular, are increasing sequences and decreasing sequences.

Arithmetic progression

Now we can define an arithmetic progression. An arithmetic progression is a simple special case of a number sequence. is a sequence in which each member, starting from the second, is equal to the previous one, to which the same number is added.

a 1 , a 2 , a 3 , . . . , a n, . . .

is an arithmetic progression if for any natural number n the condition is met:

a n +1 = a n + d,

Where d - a certain number.

Thus, the difference between the subsequent and previous terms of a given arithmetic progression is always constant:

a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.

Number d called The number is called.

To define an arithmetic progression, it is enough to indicate its first term and difference.

For example,

If a 1 = 3, d = 4 , then we find the first five terms of the sequence as follows:

a 1 =3,

a 2 = a 1 + d = 3 + 4 = 7,

a 3 = a 2 + d= 7 + 4 = 11,

a 4 = a 3 + d= 11 + 4 = 15,

a 5 = a 4 + d= 15 + 4 = 19. ◄

For an arithmetic progression with the first term a 1 and the difference d her n

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

For example,

find the thirtieth term of the arithmetic progression

1, 4, 7, 10, . . .

a 1 =1, d = 3,

a 30 = a 1 + (30 - 1)d = 1 + 29· 3 = 88. ◄

a n-1 = a 1 + (n- 2)d,

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

a n +1 = a 1 + nd,

then obviously

a n=	a n-1 + a n+1
	2

Each member of an arithmetic progression, starting from the second, is equal to the arithmetic mean of the preceding and subsequent members.

the numbers a, b and c are successive terms of some arithmetic progression if and only if one of them is equal to the arithmetic mean of the other two.

For example,

a n = 2n- 7 , is an arithmetic progression.

Let's use the above statement. We have:

a n = 2n- 7,

a n-1 = 2(n- 1) - 7 = 2n- 9,

a n+1 = 2(n+ 1) - 7 = 2n- 5.

Hence,

a n+1 + a n-1	=	2n- 5 + 2n- 9	= 2n- 7 = a n,
2		2

◄

Note that n The th term of an arithmetic progression can be found not only through a 1 , but also any previous a k

a n = a k + (n- k)d.

For example,

For a 5 can be written down

a 5 = a 1 + 4d,

a 5 = a 2 + 3d,

a 5 = a 3 + 2d,

a 5 = a 4 + d. ◄

a n = a n-k + kd,

a n = a n+k - kd,

then obviously

a n=	a n-k +a n+k
	2

any member of an arithmetic progression, starting from the second, is equal to half the sum of the equally spaced members of this arithmetic progression.

In addition, for any arithmetic progression the following equality holds:

a m + a n = a k + a l,

m + n = k + l.

For example,

in arithmetic progression

1) a 10 = 28 = (25 + 31)/2 = (a 9 + a 11 )/2;

2) 28 = a 10 = a 3 + 7d= 7 + 7 3 = 7 + 21 = 28;

3) a 10= 28 = (19 + 37)/2 = (a 7 + a 13)/2;

4) a 2 + a 12 = a 5 + a 9, because

a 2 + a 12= 4 + 34 = 38,

a 5 + a 9 = 13 + 25 = 38. ◄

S n= a 1 + a 2 + a 3 + . . .+ a n,

first n terms of an arithmetic progression is equal to the product of half the sum of the extreme terms and the number of terms:

From here, in particular, it follows that if you need to sum the terms

a k, a k +1 , . . . , a n,

then the previous formula retains its structure:

For example,

in arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, . . .

S 10 = 1 + 4 + . . . + 28 = (1 + 28) · 10/2 = 145;

10 + 13 + 16 + 19 + 22 + 25 + 28 = S 10 - S 3 = (10 + 28 ) · (10 - 4 + 1)/2 = 133. ◄

If an arithmetic progression is given, then the quantities a 1 , a n, d, n AndS n connected by two formulas:

Therefore, if the values of three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas, combined into a system of two equations with two unknowns.

An arithmetic progression is a monotonic sequence. Wherein:

If d > 0 , then it is increasing;
If d < 0 , then it is decreasing;
If d = 0 , then the sequence will be stationary.

Geometric progression

Geometric progression is a sequence in which each member, starting from the second, is equal to the previous one multiplied by the same number.

b 1 , b 2 , b 3 , . . . , b n, . . .

is a geometric progression if for any natural number n the condition is met:

b n +1 = b n · q,

Where q ≠ 0 - a certain number.

Thus, the ratio of the subsequent term of a given geometric progression to the previous one is a constant number:

b 2 / b 1 = b 3 / b 2 = . . . = b n +1 / b n = q.

Number q called denominator of geometric progression.

To define a geometric progression, it is enough to indicate its first term and denominator.

For example,

If b 1 = 1, q = -3 , then we find the first five terms of the sequence as follows:

b 1 = 1,

b 2 = b 1 · q = 1 · (-3) = -3,

b 3 = b 2 · q= -3 · (-3) = 9,

b 4 = b 3 · q= 9 · (-3) = -27,

b 5 = b 4 · q= -27 · (-3) = 81. ◄

b 1 and denominator q her n The th term can be found using the formula:

b n = b 1 · qn -1 .

For example,

find the seventh term of the geometric progression 1, 2, 4, . . .

b 1 = 1, q = 2,

b 7 = b 1 · q 6 = 1 2 6 = 64. ◄

b n-1 = b 1 · qn -2 ,

b n = b 1 · qn -1 ,

b n +1 = b 1 · qn,

then obviously

b n 2 = b n -1 · b n +1 ,

each member of the geometric progression, starting from the second, is equal to the geometric mean (proportional) of the preceding and subsequent members.

Since the converse is also true, the following statement holds:

the numbers a, b and c are successive terms of some geometric progression if and only if the square of one of them is equal to the product of the other two, that is, one of the numbers is the geometric mean of the other two.

For example,

Let us prove that the sequence given by the formula b n= -3 2 n , is a geometric progression. Let's use the above statement. We have:

b n= -3 2 n,

b n -1 = -3 2 n -1 ,

b n +1 = -3 2 n +1 .

Hence,

b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) · (-3 · 2 n +1 ) = b n -1 · b n +1 ,

which proves the desired statement. ◄

Note that n The th term of a geometric progression can be found not only through b 1 , but also any previous member b k , for which it is enough to use the formula

b n = b k · qn - k.

For example,

For b 5 can be written down

b 5 = b 1 · q 4 ,

b 5 = b 2 · q 3,

b 5 = b 3 · q 2,

b 5 = b 4 · q. ◄

b n = b k · qn - k,

b n = b n - k · q k,

then obviously

b n 2 = b n - k· b n + k

the square of any term of a geometric progression, starting from the second, is equal to the product of the terms of this progression equidistant from it.

In addition, for any geometric progression the equality is true:

b m· b n= b k· b l,

m+ n= k+ l.

For example,

in geometric progression

1) b 6 2 = 32 2 = 1024 = 16 · 64 = b 5 · b 7 ;

2) 1024 = b 11 = b 6 · q 5 = 32 · 2 5 = 1024;

3) b 6 2 = 32 2 = 1024 = 8 · 128 = b 4 · b 8 ;

4) b 2 · b 7 = b 4 · b 5 , because

b 2 · b 7 = 2 · 64 = 128,

b 4 · b 5 = 8 · 16 = 128. ◄

S n= b 1 + b 2 + b 3 + . . . + b n

first n members of a geometric progression with denominator q ≠ 0 calculated by the formula:

And when q = 1 - according to the formula

S n= nb 1

Note that if you need to sum the terms

b k, b k +1 , . . . , b n,

then the formula is used:

S n- S k -1 = b k + b k +1 + . . . + b n = b k ·	1 - qn - k +1	.
	1 - q

For example,

in geometric progression 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .

S 10 = 1 + 2 + . . . + 512 = 1 · (1 - 2 10) / (1 - 2) = 1023;

64 + 128 + 256 + 512 = S 10 - S 6 = 64 · (1 - 2 10-7+1) / (1 - 2) = 960. ◄

If a geometric progression is given, then the quantities b 1 , b n, q, n And S n connected by two formulas:

Therefore, if the values of any three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas, combined into a system of two equations with two unknowns.

For a geometric progression with the first term b 1 and denominator q the following take place properties of monotonicity :

progression is increasing if one of the following conditions is met:

b 1 > 0 And q> 1;

b 1 < 0 And 0 < q< 1;

The progression is decreasing if one of the following conditions is met:

b 1 > 0 And 0 < q< 1;

b 1 < 0 And q> 1.

If q< 0 , then the geometric progression is alternating: its terms with odd numbers have the same sign as its first term, and terms with even numbers have the opposite sign. It is clear that an alternating geometric progression is not monotonic.

Product of the first n terms of a geometric progression can be calculated using the formula:

P n= b 1 · b 2 · b 3 · . . . · b n = (b 1 · b n) n / 2 .

For example,

1 · 2 · 4 · 8 · 16 · 32 · 64 · 128 = (1 · 128) 8/2 = 128 4 = 268 435 456;

3 · 6 · 12 · 24 · 48 = (3 · 48) 5/2 = (144 1/2) 5 = 12 5 = 248 832.◄

Infinitely decreasing geometric progression

Infinitely decreasing geometric progression called an infinite geometric progression whose denominator modulus is less 1 , that is

|q| < 1 .

Note that an infinitely decreasing geometric progression may not be a decreasing sequence. It fits the occasion

1 < q< 0 .

With such a denominator, the sequence is alternating. For example,

1, - 1 / 2 , 1 / 4 , - 1 / 8 , . . . .

The sum of an infinitely decreasing geometric progression name the number to which the sum of the first ones approaches without limit n members of a progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula

S= b 1 + b 2 + b 3 + . . . =	b 1	.
	1 - q

For example,

10 + 1 + 0,1 + 0,01 + . . . = 10 / (1 - 0,1) = 11 1 / 9 ,

10 - 1 + 0,1 - 0,01 + . . . = 10 / (1 + 0,1) = 9 1 / 11 . ◄

Relationship between arithmetic and geometric progressions

Arithmetic and geometric progressions are closely related. Let's look at just two examples.

a 1 , a 2 , a 3 , . . . d Moreover, since

b a 1 , b a 2 , b a 3 , . . . b d .

For example,

1, 3, 5, . . . - arithmetic progression with difference 2 And

7 1 , 7 3 , 7 5 , . . . - geometric progression with denominator 7 2 . ◄

b 1 , b 2 , b 3 , . . . - geometric progression with denominator q Moreover, since

log a b 1, log a b 2, log a b 3, . . . - arithmetic progression with difference log aq .

For example,

2, 12, 72, . . . - geometric progression with denominator 6 And

lg 2, lg 12, lg 72, . . . - arithmetic progression with difference lg 6 . ◄

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

Solution:

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

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.

Solution:

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.

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:

Solution:

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

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

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

Number sequence

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

Number sequence
For example, for our sequence:

The assigned number is specific to only one number in the sequence. 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 number is called the th term of the sequence.

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

In our case:

Let's say we have a number sequence in which the difference between adjacent numbers is the same and equal.
For example:

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

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

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

a)
b)
c)
d)

Got it? Let's 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 term. Exists two way to find it.

1. Method

We can add the progression number to the previous value 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 term of the described arithmetic progression is equal to.

2. Method

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

For example, let’s see what the value of the th term of this arithmetic progression consists of:

In other words:

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

Did you calculate? Compare your notes with the answer:

Please note that you got exactly the same number as in the previous method, when we sequentially added the terms of the arithmetic progression to the previous value.
Let’s try to “depersonalize” this formula - let’s put it in general form and get:

Arithmetic progression equation.

Arithmetic progressions can be 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 this in practice.
We are given an arithmetic progression consisting of the following numbers: Let's check what the th number of this arithmetic progression will be if we use our formula to calculate it:

Since then:

Thus, we are convinced that the formula operates in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression yourself.

Let's compare the results:

Arithmetic progression property

Let's complicate the problem - we will derive the property of arithmetic progression.
Let's say we are given the following condition:
- arithmetic progression, find the value.
Easy, you say and start counting according to the formula you already know:

Let, ah, 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 a mistake in the calculations.
Now think about whether it is possible to solve this problem in one step using any formula? Of course yes, and that’s what we’ll try to bring out now.

Let us denote the required term of the arithmetic progression as, the formula for finding it is known to us - this is the same formula we derived at the beginning:
, Then:

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

Let's sum up the previous and subsequent terms of the progression:

It turns out that the sum of the previous and subsequent terms of the progression is the double value of the progression term located between them. In other words, to find the value of a progression term with known previous and successive values, you need to add them and divide by.

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

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

When Carl Gauss was 9 years old, a teacher, busy checking the work of students in other classes, assigned the following task in class: “Calculate the sum of all natural numbers from to (according to other sources to) inclusive.” Imagine the teacher’s surprise when one of his students (this was Karl Gauss) a minute later gave the correct answer to the task, while most of the daredevil’s classmates, after long calculations, received the wrong result...

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

Let us depict the progression given to us. Take a closer look at the highlighted numbers and try to perform various mathematical operations with them.

Have you tried it? What did you notice? Right! Their sums are equal

Now tell me, how many such pairs are there in total 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 pairs are equal, we obtain 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 difference of the progression. Try to substitute the formula of the th term into the sum formula.
What did you get?

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

How much did you get?
Gauss found that the sum of the terms is equal, and the sum of the terms. Is that what you decided?

In fact, the formula for the sum of the terms of an arithmetic progression was proven by the ancient Greek scientist Diophantus back in the 3rd century, and throughout this time, witty people made full use of the properties of the arithmetic progression.
For example, imagine Ancient Egypt and the largest construction project of that time - the construction of a pyramid... The picture 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? Calculate how many blocks are needed to build one wall if block bricks are placed at the base. I hope you won’t count while moving your finger across the monitor, 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 terms of an arithmetic progression.
Let's substitute our data into the last formulas (calculate the number of blocks in 2 ways).

Method 1.

Method 2.

And now you can calculate on the monitor: compare the obtained values with the number of blocks that are in our pyramid. Got it? Well done, you have mastered the sum of the nth terms of an arithmetic progression.
Of course, you can’t build a pyramid from 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:

Training

Tasks:

Masha is getting in shape for summer. Every day she increases the number of squats by. How many times will Masha do squats in a week if she did squats at the first training session?
What is the sum of all odd numbers contained in.
When storing logs, loggers stack them in such a way that each top layer contains one log less than the previous one. How many logs are in one masonry, if the foundation 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 do squats once a day.
First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in is half, however, let’s check this fact using the formula for finding the th term of an arithmetic progression:
Numbers do contain odd numbers.
Let's substitute the available data into the formula:
Answer: The sum of all odd numbers contained in is equal.
Let's remember the problem about pyramids. For our case, a , since each top layer is reduced by one log, then in total there are a bunch of layers, that is.
Let's substitute the data into the formula:
Answer: There are logs in the masonry.

Let's sum it up

- a number sequence in which the difference between adjacent numbers is the same and equal. It can be increasing or decreasing.
Finding formula The th term 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 is the number of numbers in progression.
The sum of the terms of an arithmetic progression can be found in two ways:

, where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

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

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

Number 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 a unique one. And we will not assign this number to any other number from this set.

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

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

It is very convenient if the th term of the sequence can be specified 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 is). Or (, difference).

nth term formula

We call a formula recurrent 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 this formula, we will have to calculate the previous nine. For example, let it. Then:

Well, is it clear now what the formula is?

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

Much more convenient 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 term is equal. What is the difference? Here's what:

(This is why it is called difference because it is equal to the difference of successive terms of the progression).

So, the formula:

Then the hundredth term is equal to:

What is the sum of all natural numbers from to?

According to legend, the great mathematician Carl Gauss, as a 9-year-old boy, calculated this amount in a few minutes. He noticed that the sum of the first and last numbers is equal, the sum of the second and penultimate is the same, the sum of the third and 3rd from the end is the same, and so on. How many such pairs are there in total? 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 subsequent number is obtained by adding to the previous number. Thus, the numbers we are interested in form an arithmetic progression with the first term and the difference.

Formula of the th term for this progression:

How many terms are there in the progression if they all have to be two-digit?

Very easy: .

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

Answer: .

Now decide for yourself:

Every day the athlete runs more meters than the previous day. How many total kilometers will he run in a week if he ran km m on the first day?
A cyclist travels more kilometers every day than the previous day. On the first day he traveled km. How many days does he need to travel to cover a kilometer? How many kilometers will he travel during the last day of his journey?
The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of a refrigerator decreased each 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: , must be found.
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 is.
Let's calculate the path traveled over the last day using the formula of the th term:
(km).
Answer:
Given: . Find: .
It couldn't be simpler:
(rub).
Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

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

Arithmetic progression can be increasing () and decreasing ().

For example:

Formula for finding the nth term of an arithmetic progression

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

Property of members of an arithmetic progression

It allows you to easily find a term of a progression if its neighboring terms are known - where is the number of numbers in the progression.

Sum of terms of an arithmetic progression

There are two ways to find the amount:

Where is the number of values.

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!

You can use our tasks (optional) and we, of course, recommend them.

In order to get better at using our tasks, you need to help extend the life of the YouClever textbook you are currently reading.

How? There are two options:

Unlock all hidden tasks in this article - 299 rub.
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Access to all hidden tasks is provided for the ENTIRE life of the site.

In conclusion...

If you don't like our tasks, find others. Just don't stop at theory.

“Understood” and “I can solve” are completely different skills. You need both.

Find problems and solve them!

Number sequence

Number sequence
For example, for our sequence:

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

In our case:

Let's say we have a number sequence in which the difference between adjacent numbers is the same and equal.
For example:

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

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

a)
b)
c)
d)

Got it? Let's 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 term. Exists two way to find it.

1. Method

We can add the progression number to the previous value 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 term of the described arithmetic progression is equal to.

2. Method

For example, let’s see what the value of the th term of this arithmetic progression consists of:

In other words:

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

Did you calculate? Compare your notes with the answer:

Arithmetic progression equation.

Arithmetic progressions can be 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:

Since then:

Thus, we are convinced that the formula operates in both decreasing and increasing arithmetic progression.
Try to find the th and th terms of this arithmetic progression yourself.

Let's compare the results:

Arithmetic progression property

Let, ah, then:

Let us denote the required term of the arithmetic progression as, the formula for finding it is known to us - this is the same formula we derived at the beginning:
, Then:

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

Let's sum up the previous and subsequent terms of the progression:

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

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

How much did you get?
Gauss found that the sum of the terms is equal, and the sum of the terms. Is that what you decided?

Method 1.

Method 2.

Training

Tasks:

Masha is getting in shape for summer. Every day she increases the number of squats by. How many times will Masha do squats in a week if she did squats at the first training session?
What is the sum of all odd numbers contained in.
When storing logs, loggers stack them in such a way that each top layer contains one log less than the previous one. How many logs are in one masonry, if the foundation 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 do squats once a day.
First odd number, last number.
Arithmetic progression difference.
The number of odd numbers in is half, however, let’s check this fact using the formula for finding the th term of an arithmetic progression:
Numbers do contain odd numbers.
Let's substitute the available data into the formula:
Answer: The sum of all odd numbers contained in is equal.
Let's remember the problem about pyramids. For our case, a , since each top layer is reduced by one log, then in total there are a bunch of layers, that is.
Let's substitute the data into the formula:
Answer: There are logs in the masonry.

Let's sum it up

- a number sequence in which the difference between adjacent numbers is the same and equal. It can be increasing or decreasing.
Finding formula The th term 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 is the number of numbers in progression.
The sum of the terms of an arithmetic progression can be found in two ways:

, where is the number of values.

ARITHMETIC PROGRESSION. AVERAGE LEVEL

Number sequence

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

Number 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 a unique one. And we will not assign this number to any other number from this set.

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

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

It is very convenient if the th term of the sequence can be specified 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 is). Or (, difference).

nth term formula

We call a formula recurrent 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 this formula, we will have to calculate the previous nine. For example, let it. Then:

Well, is it clear now what the formula is?

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

Much more convenient 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 term is equal. What is the difference? Here's what:

(This is why it is called difference because it is equal to the difference of successive terms of the progression).

So, the formula:

Then the hundredth term is equal to:

What is the sum of all natural numbers from to?

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:

Formula of the th term for this progression:

How many terms are there in the progression if they all have to be two-digit?

Very easy: .

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

Answer: .

Now decide for yourself:

Every day the athlete runs more meters than the previous day. How many total kilometers will he run in a week if he ran km m on the first day?
A cyclist travels more kilometers every day than the previous day. On the first day he traveled km. How many days does he need to travel to cover a kilometer? How many kilometers will he travel during the last day of his journey?
The price of a refrigerator in a store decreases by the same amount every year. Determine how much the price of a refrigerator decreased each 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: , must be found.
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 is.
Let's calculate the path traveled over the last day using the formula of the th term:
(km).
Answer:
Given: . Find: .
It couldn't be simpler:
(rub).
Answer:

ARITHMETIC PROGRESSION. BRIEFLY ABOUT THE MAIN THINGS

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

Arithmetic progression can be increasing () and decreasing ().

For example:

Formula for finding the nth term of an arithmetic progression

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

Property of members of an arithmetic progression

It allows you to easily find a term of a progression if its neighboring terms are known - where is the number of numbers in the progression.

Sum of terms of an arithmetic progression

There are two ways to find the amount:

Where is the number of values.

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!

You can use our tasks (optional) and we, of course, recommend them.

In order to get better at using our tasks, you need to help extend the life of the YouClever textbook you are currently reading.

How? There are two options:

Unlock all hidden tasks in this article - 299 rub.
Unlock access to all hidden tasks in all 99 articles of the textbook - 499 rub.

Yes, we have 99 such articles in our textbook and access to all tasks and all hidden texts in them can be opened immediately.

Access to all hidden tasks is provided for the ENTIRE life of the site.

In conclusion...

If you don't like our tasks, find others. Just don't stop at theory.

“Understood” and “I can solve” are completely different skills. You need both.

Find problems and solve them!