Arithmetic and geometric progressions
Theoretical information
Theoretical information
Geometric progression |
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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 |
For any natural n |
Formula nth term |
a n = a 1 + d (n – 1) |
b n = b 1 ∙ q n - 1 , b n ≠ 0 |
| Characteristic property |  |
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| Sum of the first n terms |  |
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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 arithmetic progressions, given by the formula 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 highest value n for which the inequality holds a n > -6.
Arithmetic progression problems already existed in ancient times. They appeared and demanded a solution because they had a practical need.
So, in one of the papyri Ancient Egypt", which has a 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 thought: “In an arithmetic progression, which has even number terms, the sum of the terms of the 2nd half is greater than the sum of the terms of the 1st half 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 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:
- Each term of the progression is the arithmetic mean of the previous term and the subsequent one.
- 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 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 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.
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.
Or arithmetic is a type of ordered numerical sequence, the properties of which are studied in a school algebra course. 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. The progression is given 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 first thought of this equality when he was looking for a solution to a problem given by his school teacher: 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 it?
The easiest way to answer this question is by considering the following example: let it be necessary to find the sum of terms from the m-th to the n-th. To solve the problem, you should represent the given segment from m to n of the progression as a new number series. In this view mth term a m will be 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 algebraic progression under consideration, as well as knowing what numbers in the series they occupy, you can use the formula for the sum 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 using the standard formula, then calculate the sum of the first 4 elements using the same formula, then subtract the second from the first sum.
When studying algebra in secondary school(9th grade) one of important topics is the study number sequences, which include progressions - geometric and arithmetic. In this article we will look at an arithmetic progression and examples with solutions.
What is an arithmetic progression?
To understand this, it is necessary to define the progression in question, as well as provide the basic formulas that will be used later in solving problems.
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 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:
- To determine the value of the nth term, the following formula is suitable: a n = (n-1)*d+a 1 .
- 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 member
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:
- 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.
- 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 Calculating machine will do as soon as the 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:
- S m = m * (a m + a 1) / 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.
Sum of an arithmetic progression.
The sum of an arithmetic progression is a simple thing. Both in meaning and in formula. But there are all sorts of tasks on this topic. From basic to quite solid.
First, let's understand the meaning and formula of the amount. And then we'll decide. For your own pleasure.) The meaning of the amount is as simple as a moo. To find the sum of an arithmetic progression, you just need to carefully add all its terms. If these terms are few, you can add without any formulas. But if there is a lot, or a lot... addition is annoying.) In this case, the formula comes to the rescue.
The formula for the amount is simple:
Let's figure out what kind of letters are included in the formula. This will clear things up a lot.
S n - the sum of an arithmetic progression. Addition result everyone members, with first By last. It is important. They add up exactly All members in a row, without skipping or skipping. And, precisely, starting from first. In problems like finding the sum of the third and eighth terms, or the sum of the fifth to twentieth terms, direct application of the formula will disappoint.)
a 1 - first member of the progression. Everything is clear here, it's simple first row number.
a n- last member of the progression. The last number of the series. Not a very familiar name, but when applied to the amount, it’s very suitable. Then you will see for yourself.
n - number of the last member. It is important to understand that in the formula this number coincides with the number of added terms.
Let's define the concept last member a n. Tricky question: which member will be the last one if given endless arithmetic progression?)
To answer confidently, you need to understand the elementary meaning of arithmetic progression and... read the task carefully!)
In the task of finding the sum of an arithmetic progression, the last term always appears (directly or indirectly), which should be limited. Otherwise, a final, specific amount simply doesn't exist. For the solution, it does not matter whether the progression is given: finite or infinite. It doesn’t matter how it is given: a series of numbers, or a formula for the nth term.
The most important thing is to understand that the formula works from the first term of the progression to the term with number n. Actually, the full name of the formula looks like this: the sum of the first n terms of an arithmetic progression. The number of these very first members, i.e. n, is determined solely by the task. In a task, all this valuable information is often encrypted, yes... But never mind, in the examples below we reveal these secrets.)
Examples of tasks on the sum of an arithmetic progression.
First of all, helpful information:
The main difficulty in tasks involving the sum of an arithmetic progression is correct definition elements of the formula.
The task writers encrypt these very elements with boundless imagination.) The main thing here is not to be afraid. Understanding the essence of the elements, it is enough to simply decipher them. Let's look at a few examples in detail. Let's start with a task based on a real GIA.
1. The arithmetic progression is given by the condition: a n = 2n-3.5. Find the sum of its first 10 terms.
Good job. Easy.) To determine the amount using the formula, what do we need to know? First member a 1, last term a n, yes the number of the last member n.
Where can I get the last member's number? n? Yes, right there, on condition! It says: find the sum first 10 members. Well, what number will it be with? last, tenth member?) You won’t believe it, his number is tenth!) Therefore, instead of a n We will substitute into the formula a 10, and instead n- ten. I repeat, the number of the last member coincides with the number of members.
It remains to determine a 1 And a 10. This is easily calculated using the formula for the nth term, which is given in the problem statement. Don't know how to do this? Attend the previous lesson, without this there is no way.
a 1= 2 1 - 3.5 = -1.5
a 10=2·10 - 3.5 =16.5
S n = S 10.
We have found out the meaning of all elements of the formula for the sum of an arithmetic progression. All that remains is to substitute them and count:
That's it. Answer: 75.
Another task based on the GIA. A little more complicated:
2. Given an arithmetic progression (a n), the difference of which is 3.7; a 1 =2.3. Find the sum of its first 15 terms.
We immediately write the sum formula:
This formula allows us to find the value of any term by its number. We look for a simple substitution:
a 15 = 2.3 + (15-1) 3.7 = 54.1
It remains to substitute all the elements into the formula for the sum of an arithmetic progression and calculate the answer:
Answer: 423.
By the way, if in the sum formula instead of a n We simply substitute the formula for the nth term and get:
Let us present similar ones and obtain a new formula for the sum of terms of an arithmetic progression:
 |
As you can see, the nth term is not required here a n. In some problems this formula helps a lot, yes... You can remember this formula. Or you can simply display it at the right time, like here. After all, you always need to remember the formula for the sum and the formula for the nth term.)
Now the task in the form of a short encryption):
3. Find the sum of all positive double digit numbers, multiples of three.
Wow! Neither your first member, nor your last, nor progression at all... How to live!?
You will have to think with your head and pull out all the elements of the sum of the arithmetic progression from the condition. We know what two-digit numbers are. They consist of two numbers.) What two-digit number will be first? 10, presumably.) A last thing double digit number? 99, of course! The three-digit ones will follow him...
Multiples of three... Hm... These are numbers that are divisible by three, here! Ten is not divisible by three, 11 is not divisible... 12... is divisible! So, something is emerging. You can already write down a series according to the conditions of the problem:
12, 15, 18, 21, ... 96, 99.
Will this series be an arithmetic progression? Certainly! Each term differs from the previous one by strictly three. If you add 2 or 4 to a term, say, the result, i.e. the new number is no longer divisible by 3. You can immediately determine the difference of the arithmetic progression: d = 3. It will come in handy!)
So, we can safely write down some progression parameters:
What will the number be? n last member? Anyone who thinks that 99 is fatally mistaken... The numbers always go in a row, but our members jump over three. They don't match.
There are two solutions here. One way is for the super hardworking. You can write down the progression, the entire series of numbers, and count the number of members with your finger.) The second way is for the thoughtful. You need to remember the formula for the nth term. If we apply the formula to our problem, we find that 99 is the thirtieth term of the progression. Those. n = 30.
Let's look at the formula for the sum of an arithmetic progression:
We look and rejoice.) We pulled out from the problem statement everything necessary to calculate the amount:
a 1= 12.
a 30= 99.
S n = S 30.
All that remains is elementary arithmetic. We substitute the numbers into the formula and calculate:
Answer: 1665
Another type of popular puzzle:
4. Given an arithmetic progression:
-21,5; -20; -18,5; -17; ...
Find the sum of terms from twentieth to thirty-four.
We look at the formula for the amount and... we get upset.) The formula, let me remind you, calculates the amount from the first member. And in the problem you need to calculate the sum since the twentieth... The formula won't work.
You can, of course, write out the entire progression in a series, and add terms from 20 to 34. But... it’s somehow stupid and takes a long time, right?)
There is a more elegant solution. Let's divide our series into two parts. The first part will be from the first term to the nineteenth. Second part - from twenty to thirty-four. It is clear that if we calculate the sum of the terms of the first part S 1-19, let's add it with the sum of the terms of the second part S 20-34, we get the sum of the progression from the first term to the thirty-fourth S 1-34. Like this:
S 1-19 + S 20-34 = S 1-34
From this we can see that find the sum S 20-34 can be done by simple subtraction
S 20-34 = S 1-34 - S 1-19
Both amounts on the right side are considered from the first member, i.e. the standard sum formula is quite applicable to them. Let's get started?
We extract the progression parameters from the problem statement:
d = 1.5.
a 1= -21,5.
To calculate the sums of the first 19 and first 34 terms, we will need the 19th and 34th terms. We calculate them using the formula for the nth term, as in problem 2:
a 19= -21.5 +(19-1) 1.5 = 5.5
a 34= -21.5 +(34-1) 1.5 = 28
There's nothing left. From the sum of 34 terms subtract the sum of 19 terms:
S 20-34 = S 1-34 - S 1-19 = 110.5 - (-152) = 262.5
Answer: 262.5
One important note! There is a very useful trick in solving this problem. Instead of direct calculation what you need (S 20-34), we counted something that would seem not to be needed - S 1-19. And then they determined S 20-34, discarding the unnecessary from the complete result. This kind of “feint with your ears” often saves you in wicked problems.)
In this lesson we looked at problems for which it is enough to understand the meaning of the sum of an arithmetic progression. Well, you need to know a couple of formulas.)
Practical advice:
When solving any problem involving the sum of an arithmetic progression, I recommend immediately writing out the two main formulas from this topic.
Formula for the nth term:
These formulas will immediately tell you what to look for and in what direction to think in order to solve the problem. Helps.
And now the tasks for independent solution.
5. Find the sum of all two-digit numbers that are not divisible by three.
Cool?) The hint is hidden in the note to problem 4. Well, problem 3 will help.
6. The arithmetic progression is given by the condition: a 1 = -5.5; a n+1 = a n +0.5. Find the sum of its first 24 terms.
Unusual?) This recurrence formula. You can read about it in the previous lesson. Don’t ignore the link, such problems are often found in the State Academy of Sciences.
7. Vasya saved up money for the holiday. As much as 4550 rubles! And I decided to give my favorite person (myself) a few days of happiness). Live beautifully without denying yourself anything. Spend 500 rubles on the first day, and on each subsequent day spend 50 rubles more than the previous one! Until the money runs out. How many days of happiness did Vasya have?
Is it difficult?) The additional formula from task 2 will help.
Answers (in disarray): 7, 3240, 6.
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You can get acquainted with functions and derivatives.