The sum of the first arithmetic progression. Sum of arithmetic progression

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, number sequence— function of 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 term sequences , and a natural number n — his number .

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

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 she has final number 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

Arithmetic progression 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 always constant:

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

Number d called difference of arithmetic progression.

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

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,

◄

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

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 meanings 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. In this case:

• 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:

numbers a, b and c are consecutive terms of some geometric progression if and only if the square of one of them equal to the product 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:

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 given geometric progression, 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:

Pn= 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

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 , That

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 , That

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 . ◄

Or arithmetic is a type of ordered numerical sequence, the properties of which are studied in school course algebra. This article discusses in detail the question of how to find the sum of an arithmetic progression.

What kind of progression is this?

Before moving on to the question (how to find the sum of an arithmetic progression), it is worth understanding what we are talking about.

Any sequence real numbers, which is obtained by adding (subtracting) some value from each previous number, is called an algebraic (arithmetic) progression. This definition, when translated into mathematical language, takes the form:

Here i is the serial number of the element of the row a i. Thus, knowing just one starting number, you can easily restore the entire series. The parameter d in the formula is called the progression difference.

It can be easily shown that for the series of numbers under consideration the following equality holds:

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

That is, to find the value of the nth element in order, you should add the difference d to the first element a 1 n-1 times.

What is the sum of an arithmetic progression: formula

Before giving the formula for the indicated amount, it is worth considering a simple special case. Given a progression of natural numbers from 1 to 10, you need to find their sum. Since there are few terms in the progression (10), it is possible to solve the problem head-on, that is, sum all the elements in order.

S 10 = 1+2+3+4+5+6+7+8+9+10 = 55.

One thing worth considering interesting thing: since each term differs from the next one by the same value d = 1, then the pairwise summation of the first with the tenth, the second with the ninth, and so on will give the same result. Really:

11 = 1+10 = 2+9 = 3+8 = 4+7 = 5+6.

As you can see, there are only 5 of these sums, that is, exactly two times less than the number of elements of the series. Then multiplying the number of sums (5) by the result of each sum (11), you will arrive at the result obtained in the first example.

If we generalize these arguments, we can write the following expression:

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

This expression shows that it is not at all necessary to sum all the elements in a row; it is enough to know the value of the first a 1 and the last a n , as well as total number n terms.

It is believed that Gauss was the first to think of this equality when he was looking for a solution to a given problem. school teacher task: sum the first 100 integers.

Sum of elements from m to n: formula

The formula given in the previous paragraph answers the question of how to find the sum of an arithmetic progression (the first elements), but often in problems it is necessary to sum a series of numbers in the middle of the progression. How to do this?

The easiest way to answer this question is by considering next example: let it be necessary to find the sum of terms from m-th to n-th. To solve the problem, you should present the given segment from m to n of the progression in the form of a new number series. In this m-th representation the term a m will be the first, and a n will be numbered n-(m-1). In this case, applying the standard formula for the sum, the following expression will be obtained:

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

Example of using formulas

Knowing how to find the sum of an arithmetic progression, it is worth considering a simple example of using the above formulas.

Below is a numerical sequence, you should find the sum of its terms, starting from the 5th and ending with the 12th:

The given numbers indicate that the difference d is equal to 3. Using the expression for the nth element, you can find the values ​​of the 5th and 12th terms of the progression. It turns out:

a 5 = a 1 + d * 4 = -4 + 3 * 4 = 8;

a 12 = a 1 + d * 11 = -4 + 3 * 11 = 29.

Knowing the values ​​of the numbers at the ends of the given algebraic progression, and also knowing what numbers in the row they occupy, you can use the formula for the amount obtained in the previous paragraph. It will turn out:

S 5 12 = (12 - 5 + 1) * (8 + 29) / 2 = 148.

It is worth noting that this value could be obtained differently: first find the sum of the first 12 elements by standard formula, then calculate the sum of the first 4 elements using the same formula, then subtract the second from the first sum.

Problems on arithmetic progression existed already in ancient times. They appeared and demanded a solution because they had a practical need.

So, in one of the papyri Ancient Egypt having mathematical content, - the Rhind papyrus (19th century BC) - contains the following task: divide ten measures of bread among ten people, provided that the difference between each of them is one eighth of the measure.”

And in the mathematical works of the ancient Greeks there are elegant theorems related to arithmetic progression. Thus, Hypsicles of Alexandria (2nd century, which amounted to a lot interesting tasks and who added the fourteenth book to Euclid’s Elements, formulated the thought: “In an arithmetic progression that has an even number of terms, the sum of the terms of the 2nd half is greater than the sum of the terms of the 1st by the square of 1/2 the number of terms.”

The sequence is denoted by an. The numbers of a sequence are called its members and are usually designated by letters with indices that indicate the serial number of this member (a1, a2, a3 ... read: “a 1st”, “a 2nd”, “a 3rd” and so on ).

The sequence can be infinite or finite.

What is an arithmetic progression? By it we mean the one obtained by adding the previous term (n) with the same number d, which is the difference of the progression.

If d<0, то мы имеем убывающую прогрессию. Если d>0, then this progression is considered increasing.

An arithmetic progression is called finite if only its first few terms are taken into account. At very large quantities members it's already endless progression.

Any arithmetic progression is defined by the following formula:

an =kn+b, while b and k are some numbers.

The opposite statement is absolutely true: if a sequence is given by a similar formula, then it is exactly an arithmetic progression that has the properties:

1. Each term of the progression is the arithmetic mean of the previous term and the subsequent one.
2. Converse: if, starting from the 2nd, each term is the arithmetic mean of the previous term and the subsequent one, i.e. if the condition is met, then this sequence is an arithmetic progression. This equality is also a sign of progression, which is why it is usually called characteristic property progression.
   In the same way, the theorem that reflects this property is true: a sequence is an arithmetic progression only if this equality is true for any of the terms of the sequence, starting with the 2nd.

The characteristic property for any four numbers of an arithmetic progression can be expressed by the formula an + am = ak + al, if n + m = k + l (m, n, k are progression numbers).

In an arithmetic progression, any necessary (Nth) term can be found using the following formula:

For example: the first term (a1) in an arithmetic progression is given and equal to three, and the difference (d) is equal to four. You need to find the forty-fifth term of this progression. a45 = 1+4(45-1)=177

The formula an = ak + d(n - k) allows us to determine nth term an arithmetic progression through any of its kth terms, provided that it is known.

The sum of the terms of an arithmetic progression (meaning the 1st n terms finite progression) is calculated as follows:

Sn = (a1+an) n/2.

If the 1st term is also known, then another formula is convenient for calculation:

Sn = ((2a1+d(n-1))/2)*n.

The sum of an arithmetic progression that contains n terms is calculated as follows:

The choice of formulas for calculations depends on the conditions of the problems and the initial data.

Natural series of any numbers, such as 1,2,3,...,n,...- simplest example arithmetic progression.

In addition to the arithmetic progression, there is also a geometric progression, which has its own properties and characteristics.

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.

The number sequence is 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 sequence, unlike 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 one by one, 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 Latin word recurro- come back.

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

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

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

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

For example, 2; 5; 8; 11;...

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 picture.

We see that

, and at the same time

Adding these two equalities, we get:

.

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

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

Moreover, since

, and at the same time

, That

, and therefore

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

Formula of the th term.

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

and finally

We got formula of the nth term.

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

The sum of n terms of an arithmetic progression.

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

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

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

Let's add in pairs:

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

We get:

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

Let's consider solving arithmetic progression problems.

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

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

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

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

a) Find 31 terms of the progression.

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

A) We see that ;

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

In general

In our case , That's why

We get:

b) Suppose the number 41 is a member of the sequence. Let's find his number. To do this, let's solve the equation:

We got natural value n, therefore, yes, the number 41 is a member of the progression. If the found value of n were not a natural number, then we would answer that the number 41 is NOT a member of the progression.

3 . a) Between numbers 2 and 8, insert 4 numbers so that they, together with these numbers, form an arithmetic progression.

b) Find the sum of the terms of the resulting progression.

A) Let's insert four numbers between numbers 2 and 8:

We got an arithmetic progression with 6 members.

Let's find the difference of this progression. To do this, we use the formula for the nth term:

Now it's easy to find the meanings of the numbers:

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

b)

Answer: a) yes; b) 30

4. The truck transports a load of crushed stone weighing 240 tons, increasing the transportation rate by the same number of tons every day. It is known that 2 tons of crushed stone were transported on the first day. Determine how many tons of crushed stone were transported on the twelfth day if all the work was completed in 15 days.

According to the condition of the problem, the amount of crushed stone that the truck transports increases by the same number every day. Therefore, we are dealing with an arithmetic progression.

Let us formulate this problem in terms of an arithmetic progression.

During the first day, 2 tons of crushed stone were transported: a_1=2.

All work was completed in 15 days: .

The truck is transporting a batch of crushed stone weighing 240 tons:

We need to find.

First, let's find the progression difference. Let's use the formula for the sum of n terms of a progression.

In our case:

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

First, a couple of examples. Let's look at several sets of numbers:

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

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

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

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

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

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

And a couple at once important comments. Firstly, progression is only considered ordered sequence of numbers: they are allowed to be read strictly in the order in which they are written - and nothing else. Numbers cannot be rearranged or swapped.

Secondly, the sequence itself can be either finite or infinite. For example, the set (1; 2; 3) is obviously a finite arithmetic progression. But if you write something in the spirit (1; 2; 3; 4; ...) - this is already an infinite progression. The ellipsis after the four seems to hint that there are quite a few more numbers to come. Infinitely many, for example. :)

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

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

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

Definition. An arithmetic progression is called:

1. increasing if each next element is greater than the previous one;
2. decreasing if, on the contrary, each subsequent element is less than the previous one.

In addition, there are so-called “stationary” sequences - they consist of the same repeating number. For example, (3; 3; 3; ...).

Only one question remains: how to distinguish an increasing progression from a decreasing one? Fortunately, everything here depends only on the sign of the number $d$, i.e. progression differences:

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

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

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

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

Progression terms and recurrence formula

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

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

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

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

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

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

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

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

However, I suggest you practice a little.

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

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

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

Answer: (8; 3; −2)

That's it! Please note: our progression is decreasing.

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

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

Solution. Let us write the problem condition in familiar terms:

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

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

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

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

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

That's how easy it is to find the progression difference! All that remains is to substitute the found number into any of the equations of the system. For example, in the first:

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

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

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

Ready! The problem is solved.

Answer: (−34; −35; −36)

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

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

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

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

Solution. Since $((a)_(5))=8.4$, $((a)_(10))=14.4$, and we need to find $((a)_(15))$, we note following:

\[\begin(align) & ((a)_(15))-((a)_(10))=5d; \\ & ((a)_(10))-((a)_(5))=5d. \\ \end(align)\]

But by condition $((a)_(10))-((a)_(5))=14.4-8.4=6$, therefore $5d=6$, from which we have:

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

Answer: 20.4

That's it! We didn’t need to create any systems of equations and calculate the first term and the difference - everything was solved in just a couple of lines.

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

At the same time, it is not always possible to find this moment “head-on” by sequentially going through the elements. Often, problems are written in such a way that without knowing the formulas, the calculations would take several sheets of paper—we would simply fall asleep while we found the answer. Therefore, let's try to solve these problems in a faster way.

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

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

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

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

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

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

Task No. 5. In arithmetic progression $(()_(5))=-150,(()_(6))=-147$. Find the number of the first positive term of this progression.

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

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

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

Now we proceed by analogy with previous task. Let's find out at what point in our sequence positive numbers will appear:

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

Minimal integer solution of this inequality- number 56.

Please note: in the last task it all came down to strict inequality, so the option $n=55$ will not suit us.

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

Arithmetic mean and equal indentations

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

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

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

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

However, these equalities can be rewritten differently:

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

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

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

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

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

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

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

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

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

Solution. Let us again express the middle term through the arithmetic mean of neighboring terms:

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

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

Answer: 1; 6.

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

Let's say in problem No. 6 we received answers −3 and 2. How can we check that these answers are correct? Let's just plug them into the original condition and see what happens. Let me remind you that we have three numbers ($-6(()^(2))$, $+1$ and $14+4(()^(2))$), which must form an arithmetic progression. Let's substitute $x=-3$:

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

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

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

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

Overall, deciding latest tasks, we came across another one interesting fact, which also needs to be remembered:

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

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

Grouping and summing elements

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

Let's try to express the “left tail” through $((a)_(n))$ and $d$, and the “right tail” through $((a)_(k))$ and $d$. It's very simple:

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

Now note that the following amounts are equal:

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

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

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

Task No. 8. Determine the difference of an arithmetic progression in which the first term is 66, and the product of the second and twelfth terms is the smallest possible.

Solution. Let's write down everything we know:

\[\begin(align) & ((a)_(1))=66; \\&d=? \\ & ((a)_(2))\cdot ((a)_(12))=\min . \end(align)\]

So, we do not know the progression difference $d$. Actually, the entire solution will be built around the difference, since the product $((a)_(2))\cdot ((a)_(12))$ can be rewritten as follows:

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

For those in the tank: I took it out common multiplier 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 expand the brackets, we get:

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

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

schedule quadratic function- parabola

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

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

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

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

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

Answer: −36

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

Solution. Essentially, we need to make a sequence of five numbers, with the first and last number is already known. Let's denote the missing numbers by the variables $x$, $y$ and $z$:

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

Note that the number $y$ is the “middle” of our sequence - it is equidistant from the numbers $x$ and $z$, and from the numbers $-\frac(1)(2)$ and $-\frac(1)( 6)$. And if from the numbers $x$ and $z$ we are in at the moment we can’t get $y$, then the situation is different with the ends of the progression. Let's remember the arithmetic mean:

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

Using similar reasoning, we find the remaining number:

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

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

Task No. 10. Between the numbers 2 and 42, insert several numbers that, together with these numbers, form an arithmetic progression, if you know that the sum of the first, second and last of the inserted numbers is 56.

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

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

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

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

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

But then the expression written above can be rewritten as follows:

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

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

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

All that remains is to find the remaining terms:

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

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

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

Word problems with progressions

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

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

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

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

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

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

Therefore, 202 parts will be produced in November.

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

Solution. Everything is the same:

$\begin(align) & ((a)_(1))=216;\quad d=4; \\ & ((a)_(n))=216+\left(n-1 \right)\cdot 4. \\ \end(align)$

December is the last, 12th month of the year, so we are looking for $((a)_(12))$:

\[((a)_(12))=216+11\cdot 4=260\]

This is the answer - 260 books will be bound in December.

Well, if you have read this far, I hasten to congratulate you: you have successfully completed the “young fighter’s course” in arithmetic progressions. You can safely move on to the next lesson, where we will study the formula for the sum of progression, as well as important and very useful consequences from her.