Arithmetic progression d. How to find the sum of an arithmetic progression: formulas and an example of their use

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

Thus, one of the papyri of Ancient Egypt that has 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, who compiled many interesting problems and added the fourteenth book to Euclid’s Elements), formulated the idea: “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 on the square 1/ 2 numbers of members."

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. With a very large number of members, this is already an 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 a characteristic property of 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 you to determine the nth term of 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 first n terms of a 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.

The natural series of any numbers, such as 1,2,3,...,n,..., is the simplest example of an arithmetic progression.

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

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

What is an arithmetic progression?

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

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

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

Important Formulas

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

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

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

Example #1: finding an unknown term

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

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

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

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

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

Example #2: progression difference

Now let’s complicate the task a little, let’s give an example of how

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

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

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

Example No. 3: drawing up a progression

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

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

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

Example No. 4: first term of progression

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

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

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

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

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

Example No. 5: amount

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

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

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

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

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

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

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

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

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

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

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

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

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

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

I. V. Yakovlev | Mathematics materials | MathUs.ru

Arithmetic progression

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

Subsequence

Imagine a device on the screen of which certain numbers are displayed one after another. Let's say 2; 7; 13; 1; 6; 0; 3; : : : This set of numbers is precisely an example of a sequence.

Definition. A number sequence is a set of numbers in which each number can be assigned a unique number (that is, associated with a single natural number)1. The number n is called the nth term of the sequence.

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

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

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

Arithmetic progression: basic definitions

Now we are ready to define an arithmetic progression.

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

For example, sequence 2; 5; 8; eleven; : : : is an arithmetic progression with first term 2 and difference 3. Sequence 7; 2; 3; 8; : : : is an arithmetic progression with first term 7 and difference 5. Sequence 3; 3; 3; : : : is an arithmetic progression with a difference equal to zero.

Equivalent definition: the sequence an is called an arithmetic progression if the difference an+1 an is a constant value (independent of n).

An arithmetic progression is called increasing if its difference is positive, and decreasing if its difference is negative.

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

By default, sequences are considered infinite, that is, containing an infinite number of numbers. But no one bothers us to consider finite sequences; in fact, any finite set of numbers can be called a finite sequence. For example, the ending sequence is 1; 2; 3; 4; 5 consists of five numbers.

Formula for the nth term of an arithmetic progression

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

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

Problem 1. In arithmetic progression 2; 5; 8; eleven; : : : find the formula for the nth term and calculate the hundredth term.

Solution. According to formula (1) we have:

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

a100 = 3 100 1 = 299:

Property and sign of arithmetic progression

Property of arithmetic progression. In arithmetic progression an for any

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

which is what was required.

More generally, the arithmetic progression an satisfies the equality

a n = a n k + a n+k

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

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

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

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

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

From this we can see that the difference an+1 an does not depend on n, and this precisely means that the sequence an is an arithmetic progression.

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

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

Problem 2. (MSU, Faculty of Economics, 2007) Three numbers 8x, 3 x2 and 4 in the indicated order form a decreasing arithmetic progression. Find x and indicate the difference of this progression.

Solution. By the property of arithmetic progression we have:

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

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

Answer: x = 1, the difference is 6.

Sum of the first n terms of an arithmetic progression

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

Little Gauss's idea was as follows. Let

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

Let's write this amount in reverse order:

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

and add these two formulas:

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

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

2S = 101 100 = 10100;

We use this idea to derive the sum formula

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

A useful modification of formula (3) is obtained if we substitute the formula of the nth term an = a1 + (n 1)d into it:

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

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

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

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

an 6 999; 91 + 13n 6 999;

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

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

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

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

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

However, \(d\) can also be a negative number. For example, in arithmetic progression \(16\); \(10\); \(4\); \(-2\); \(-8\)... the progression difference \(d\) is equal to minus six.

And in this case, each next element will be smaller than the previous one. These progressions are called decreasing.

Arithmetic progression notation

Progression is indicated by a small Latin letter.

Numbers that form a progression are called members(or elements).

They are denoted by the same letter as an arithmetic progression, but with a numerical index equal to the number of the element in order.

For example, the arithmetic progression \(a_n = \left\( 2; 5; 8; 11; 14…\right\)\) consists of the elements \(a_1=2\); \(a_2=5\); \(a_3=8\) and so on.

In other words, for the progression \(a_n = \left\(2; 5; 8; 11; 14…\right\)\)

Solving arithmetic progression problems

In principle, the information presented above is already enough to solve almost any arithmetic progression problem (including those offered at the OGE).

Example (OGE). The arithmetic progression is specified by the conditions \(b_1=7; d=4\). Find \(b_5\).
Solution:

Answer: \(b_5=23\)

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

Answer: \(-3\)

Example (OGE). Given several consecutive elements of an arithmetic progression: \(…5; x; 10; 12.5...\) Find the value of the element designated by the letter \(x\).
Solution:

Answer: \(7,5\).

Example (OGE). The arithmetic progression is defined by the following conditions: \(a_1=-11\); \(a_(n+1)=a_n+5\) Find the sum of the first six terms of this progression.
Solution:

Answer: \(S_6=9\).

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

Answer: \(d=7\).

Important formulas for arithmetic progression

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

However, sometimes there are situations when deciding “head-on” is very inconvenient. For example, imagine that in the very first example we need to find not the fifth element \(b_5\), but the three hundred and eighty-sixth \(b_(386)\). Should we add four \(385\) times? Or imagine that in the penultimate example you need to find the sum of the first seventy-three elements. You'll be tired of counting...

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

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

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

Example. The arithmetic progression is specified by the conditions: \(b_1=-159\); \(d=8.2\). Find \(b_(246)\).
Solution:

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

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

\(a_n\) – the last summed term;

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

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

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

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

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

Example. Find the sum of the first \(33\)-ex terms of the arithmetic progression: \(17\); \(15.5\); \(14\)…
Solution:

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

More complex arithmetic progression problems

Now you have all the information you need to solve almost any arithmetic progression problem. Let’s finish the topic by considering problems in which you not only need to apply formulas, but also think a little (in mathematics this can be useful ☺)

Example (OGE). Find the sum of all negative terms of the progression: \(-19.3\); \(-19\); \(-18.7\)…
Solution:

Answer: \(S_(65)=-630.5\).

Example (OGE). The arithmetic progression is specified by the conditions: \(a_1=-33\); \(a_(n+1)=a_n+4\). Find the sum from the \(26\)th to the \(42\) element inclusive.
Solution:

Answer: \(S=1683\).

For arithmetic progression, there are several more formulas that we did not consider in this article due to their low practical usefulness. However, you can easily find them.

Arithmetic and geometric progressions

Theoretical information

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

Examples of tasks with comments

Exercise 1

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

According to the formula of the nth term:

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

By condition:

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

It is necessary to find the difference of progressions:

d = a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = - 48.

Answer : a 22 = -48.

Task 2

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

1st method (using the n-term formula)

According to the formula for the nth term of a geometric progression:

b 5 = b 1 ∙ q 5 - 1 = b 1 ∙ q 4.

Because b 1 = -3,

2nd method (using recurrent formula)

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

b 3 = 6 ∙ (-2) = -12;

b 4 = -12 ∙ (-2) = 24;

b 5 = 24 ∙ (-2) = -48.

Answer : b 5 = -48.

Task 3

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

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

Therefore:

.

Let's substitute the data into the formula:

Answer: 95.

Task 4

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

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

.

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

By condition, the formula for the nth term of the original progression is known ( a n) a n= 3n - 4. You can find immediately and a 1, And a 16 without finding d. Therefore, we will use the first formula.

Answer: 368.

Task 5

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

According to the formula of the nth term:

a 22 = a 1 + d (22 – 1) = a 1+ 21d.

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

d = a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = -48.

Answer : a 22 = -48.

Task 6

Several consecutive terms of the geometric progression are written:

Find the term of the progression indicated by x.

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

Substituting the found values ​​into the formula, we get:

.

Answer : .

Task 7

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

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

.

Answer: 4.

Task 8

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