Arithmetic progression formulas examples solutions. Arithmetic and geometric progressions
For example, the sequence \(2\); \(5\); \(8\); \(11\); \(14\)... 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 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, 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:
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We are given the first elements of the sequence and know that it is an arithmetic progression. That is, each element differs from its neighbor by the same number. Let's find out which one by subtracting the previous one from the next element: \(d=49-62=-13\). |
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Now we can restore our progression to the (first negative) element we need. |
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Ready. You can write an answer. |
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:
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To find \(x\), we need to know how much the next element differs from the previous one, in other words, the progression difference. Let's find it from two known neighboring elements: \(d=12.5-10=2.5\). |
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And now we can easily find what we are looking for: \(x=5+2.5=7.5\). |
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Ready. You can write an answer. |
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:
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We need to find the sum of the first six terms of the progression. But we don’t know their meanings; we are given only the first element. Therefore, we first calculate the values one by one, using what is given to us: \(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\) |
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\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\) |
The required amount has been found. |
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:
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\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\) |
To calculate the sum of the first twenty-five terms, we need to know the value of the first and twenty-fifth terms. |
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\(n=1;\) \(a_1=3.4·1-0.6=2.8\) |
Now let's find the twenty-fifth term by substituting twenty-five instead of \(n\). |
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\(n=25;\) \(a_(25)=3.4·25-0.6=84.4\) |
Well, now we can easily calculate the required amount. |
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\(S_(25)=\)\(\frac(a_1+a_(25))(2)\) \(\cdot 25=\) |
The answer is ready. |
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:
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\(S_n=\)\(\frac(2a_1+(n-1)d)(2)\) \(\cdot n\) |
The task is very similar to the previous one. We begin to solve the same thing: first we find \(d\). |
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\(d=a_2-a_1=-19-(-19.3)=0.3\) |
Now I would like to substitute \(d\) into the formula for the sum... and here a small nuance emerges - we do not know \(n\). In other words, we don’t know how many terms will need to be added. How to find out? Let's think. We will stop adding elements when we reach the first positive element. That is, you need to find out the number of this element. How? Let's write down the formula for calculating any element of an arithmetic progression: \(a_n=a_1+(n-1)d\) for our case. |
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\(a_n=a_1+(n-1)d\) |
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\(a_n=-19.3+(n-1)·0.3\) |
We need \(a_n\) to become greater than zero. Let's find out at what \(n\) this will happen. |
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\(-19.3+(n-1)·0.3>0\) |
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\((n-1)·0.3>19.3\) \(|:0.3\) |
We divide both sides of the inequality by \(0.3\). |
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\(n-1>\)\(\frac(19.3)(0.3)\) |
We transfer minus one, not forgetting to change the signs |
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\(n>\)\(\frac(19.3)(0.3)\) \(+1\) |
Let's calculate... |
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\(n>65,333…\) |
...and it turns out that the first positive element will have the number \(66\). Accordingly, the last negative one has \(n=65\). Just in case, let's check this. |
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\(n=65;\) \(a_(65)=-19.3+(65-1)·0.3=-0.1\) |
So we need to add the first \(65\) elements. |
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\(S_(65)=\) \(\frac(2 \cdot (-19.3)+(65-1)0.3)(2)\)\(\cdot 65\) |
The answer is ready. |
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:
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\(a_1=-33;\) \(a_(n+1)=a_n+4\) |
In this problem you also need to find the sum of elements, but starting not from the first, but from the \(26\)th. For such a case we do not have a formula. How to decide? |
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For our progression \(a_1=-33\), and the difference \(d=4\) (after all, it is the four that we add to the previous element to find the next one). Knowing this, we find the sum of the first \(42\)-y elements. |
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\(S_(42)=\) \(\frac(2 \cdot (-33)+(42-1)4)(2)\)\(\cdot 42=\) |
Now the sum of the first \(25\) elements. |
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\(S_(25)=\) \(\frac(2 \cdot (-33)+(25-1)4)(2)\)\(\cdot 25=\) |
And finally, we calculate the answer. |
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\(S=S_(42)-S_(25)=2058-375=1683\) |
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.
Before we start deciding arithmetic progression problems, let's consider what a number sequence is, since an arithmetic progression is a special case of a number sequence.
A number sequence is a number set, each element of which has its own serial number. The elements of this set are called members of the sequence. The serial number of a sequence element is indicated by an index:
The first element of the sequence;
The fifth element of the sequence;
- the “nth” element of the sequence, i.e. element "standing in queue" at number n.
There is a relationship between the value of a sequence element and its sequence number. Therefore, we can consider a sequence as a function whose argument is the ordinal number of the element of the sequence. In other words, we can say that the sequence is a function of the natural argument:
The sequence can be set in three ways:
1 . The sequence can be specified using a table. In this case, we simply set the value of each member of the sequence.
For example, Someone decided to take up personal time management, and to begin with, count how much time he spends on VKontakte during the week. By recording the time in the table, he will receive a sequence consisting of seven elements:
The first line of the table indicates the number of the day of the week, the second - the time in minutes. We see that, that is, on Monday Someone spent 125 minutes on VKontakte, that is, on Thursday - 248 minutes, and, that is, on Friday only 15.
2 . The sequence can be specified using the nth term formula.
In this case, the dependence of the value of a sequence element on its number is expressed directly in the form of a formula.
For example, if , then
To find the value of a sequence element with a given number, we substitute the element number into the formula of the nth term.
We do the same thing if we need to find the value of a function if the value of the argument is known. We substitute the value of the argument into the function equation:
If, for example, 
, That
Let me note once again that in 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 the Latin word recurro- come back.
Now we can define an arithmetic progression. An arithmetic progression is a simple special case of a number sequence.
Arithmetic progression is a numerical sequence, each member of which, starting from the second, is equal to the previous one added to the same number.
The number is called difference of arithmetic progression. The difference of an arithmetic progression can be positive, negative, or equal to zero.
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 
Online calculator.
Solving an arithmetic progression.
Given: a n , d, n
Find: a 1
This math program finds \(a_1\) of an arithmetic progression based on user-specified numbers \(a_n, d\) and \(n\).
The numbers \(a_n\) and \(d\) can be specified not only as integers, but also as fractions. Moreover, fractional number can be entered as a decimal fraction (\(2.5\)) and as common fraction(\(-5\frac(2)(7)\)).
The program not only gives the answer to the problem, but also displays the process of finding a solution.
This online calculator may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.
In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.
If you are not familiar with the rules for entering numbers, we recommend that you familiarize yourself with them.
Rules for entering numbers
The numbers \(a_n\) and \(d\) can be specified not only as integers, but also as fractions.
The number \(n\) can only be a positive integer.
Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a period or a comma.
For example, you can enter decimals so 2.5 or so 2.5
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
Input:
Result: \(-\frac(2)(3)\)
Whole part separated from the fraction by an ampersand: &
Input:
Result: \(-1\frac(2)(3)\)
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A little theory.
Number sequence
In everyday practice, numbering of various objects is often used to indicate the order in which they are arranged. For example, the houses on each street are numbered. In the library, reader's subscriptions are numbered and then arranged in the order of assigned numbers in special card files.
In a savings bank, using the depositor’s personal account number, you can easily find this account and see what deposit is on it. Let account No. 1 contain a deposit of a1 rubles, account No. 2 contain a deposit of a2 rubles, etc. It turns out number sequence
a 1 , a 2 , a 3 , ..., a N
where N is the number of all accounts. Here, each natural number n from 1 to N is associated with a number a n.
Also studied in mathematics infinite number sequences:
a 1 , a 2 , a 3 , ..., a n , ... .
The number a 1 is called first term of the sequence, number a 2 - second term of the sequence, number a 3 - third term of the sequence etc.
The number a n is called nth (nth) member of the sequence, and the natural number n is its number.
For example, in a sequence of squares natural numbers 1, 4, 9, 16, 25, ..., n 2, (n + 1) 2, ... and 1 = 1 is the first term of the sequence; and n = n 2 is nth term sequences; a n+1 = (n + 1) 2 is the (n + 1)th (n plus first) term of the sequence. Often a sequence can be specified by the formula of its nth term. For example, the formula \(a_n=\frac(1)(n), \; n \in \mathbb(N) \) defines the sequence \(1, \; \frac(1)(2) , \; \frac( 1)(3) , \; \frac(1)(4) , \dots,\frac(1)(n) , \dots \)
Arithmetic progression
The length of the year is approximately 365 days. A more accurate value is \(365\frac(1)(4)\) days, so every four years an error of one day accumulates.
To account for this error, a day is added to every fourth year, and the extended year is called a leap year.
For example, in the third millennium leap years are the years 2004, 2008, 2012, 2016, ... .
In this sequence, each member, starting from the second, is equal to the previous one, added to the same number 4. Such sequences are called arithmetic progressions.
Definition.
The number sequence a 1, a 2, a 3, ..., a n, ... is called arithmetic progression, if for all natural n the equality
\(a_(n+1) = a_n+d, \)
where d is some number.
From this formula it follows that a n+1 - a n = d. The number d is called the difference arithmetic progression.
By definition of an arithmetic progression we have:
\(a_(n+1)=a_n+d, \quad a_(n-1)=a_n-d, \)
where
\(a_n= \frac(a_(n-1) +a_(n+1))(2) \), where \(n>1 \)
Thus, each term of an arithmetic progression, starting from the second, is equal to the arithmetic mean of its two adjacent terms. This explains the name "arithmetic" progression.
Note that if a 1 and d are given, then the remaining terms of the arithmetic progression can be calculated using the recurrent formula a n+1 = a n + d. In this way it is not difficult to calculate the first few terms of the progression, however, for example, a 100 will already require a lot of calculations. Typically, the nth term formula is used for this. By definition of arithmetic progression
\(a_2=a_1+d, \)
\(a_3=a_2+d=a_1+2d, \)
\(a_4=a_3+d=a_1+3d \)
etc.
At all,
\(a_n=a_1+(n-1)d, \)
because nth term of an arithmetic progression is obtained from the first term by adding (n-1) times the number d.
This formula is called formula for the nth term of an arithmetic progression.
Sum of the first n terms of an arithmetic progression
Find the sum of all natural numbers from 1 to 100.
Let's write this amount in two ways:
S = l + 2 + 3 + ... + 99 + 100,
S = 100 + 99 + 98 + ... + 2 + 1.
Let's add these equalities term by term:
2S = 101 + 101 + 101 + ... + 101 + 101.
This sum has 100 terms
Therefore, 2S = 101 * 100, hence S = 101 * 50 = 5050.
Let us now consider an arbitrary arithmetic progression
a 1 , a 2 , a 3 , ..., a n , ...
Let S n be the sum of the first n terms of this progression:
S n = a 1 , a 2 , a 3 , ..., a n
Then the sum of the first n terms of an arithmetic progression is equal to
\(S_n = n \cdot \frac(a_1+a_n)(2) \)
Since \(a_n=a_1+(n-1)d\), then replacing a n in this formula we get another formula for finding sum of the first n terms of an arithmetic progression:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)
Some people treat the word “progression” with caution, as a very complex term from the branches of higher mathematics. Meanwhile, the simplest arithmetic progression is the work of the taxi meter (where they still exist). And understanding the essence (and in mathematics there is nothing more important than “getting the essence”) of an arithmetic sequence is not so difficult, having analyzed a few elementary concepts.
Mathematical number sequence
A numerical sequence is usually called a series of numbers, each of which has its own number.
a 1 is the first member of the sequence;
and 2 is the second term of the sequence;
and 7 is the seventh member of the sequence;
and n is the nth member of the sequence;
However, not any arbitrary set of numbers and numbers interests us. We will focus our attention on a numerical sequence in which the value of the nth term is related to its ordinal number by a relationship that can be clearly formulated mathematically. In other words: the numerical value of the nth number is some function of n.
a is the value of a member of a numerical sequence;
n is its serial number;
f(n) is a function, where the ordinal number in the numerical sequence n is the argument.
Definition
An arithmetic progression is usually called a numerical sequence in which each subsequent term is greater (less) than the previous one by the same number. The formula for the nth term of an arithmetic sequence is as follows:
a n - the value of the current member of the arithmetic progression;
a n+1 - formula of the next number;
d - difference (certain number).
It is easy to determine that if the difference is positive (d>0), then each subsequent member of the series under consideration will be greater than the previous one and such an arithmetic progression will be increasing.
In the graph below it is easy to see why the number sequence is called “increasing”.
In cases where the difference is negative (d<0), каждый последующий член по понятным причинам будет меньше предыдущего, график прогрессии станет «уходить» вниз, арифметическая прогрессия, соответственно, будет именоваться убывающей.
Specified member value
Sometimes it is necessary to determine the value of any arbitrary term a n of an arithmetic progression. This can be done by sequentially calculating the values of all members of the arithmetic progression, starting from the first to the desired one. However, this path is not always acceptable if, for example, it is necessary to find the value of the five thousandth or eight millionth term. Traditional calculations will take a lot of time. However, a specific arithmetic progression can be studied using certain formulas. There is also a formula for the nth term: the value of any term of an arithmetic progression can be determined as the sum of the first term of the progression with the difference of the progression, multiplied by the number of the desired term, reduced by one.
The formula is universal for increasing and decreasing progression.
An example of calculating the value of a given term
Let us solve the following problem of finding the value of the nth term of an arithmetic progression.
Condition: there is an arithmetic progression with parameters:
The first term of the sequence is 3;
The difference in the number series is 1.2.
Task: you need to find the value of 214 terms
Solution: to determine the value of a given term, we use the formula:
a(n) = a1 + d(n-1)
Substituting the data from the problem statement into the expression, we have:
a(214) = a1 + d(n-1)
a(214) = 3 + 1.2 (214-1) = 258.6
Answer: The 214th term of the sequence is equal to 258.6.
The advantages of this method of calculation are obvious - the entire solution takes no more than 2 lines.
Sum of a given number of terms
Very often, in a given arithmetic series, it is necessary to determine the sum of the values of some of its segments. To do this, there is also no need to calculate the values of each term and then add them up. This method is applicable if the number of terms whose sum needs to be found is small. In other cases, it is more convenient to use the following formula.
The sum of the terms of an arithmetic progression from 1 to n is equal to the sum of the first and nth terms, multiplied by the number of the term n and divided by two. If in the formula the value of the nth term is replaced by the expression from the previous paragraph of the article, we get:
Calculation example
For example, let’s solve a problem with the following conditions:
The first term of the sequence is zero;
The difference is 0.5.
The problem requires determining the sum of the terms of the series from 56 to 101.
Solution. Let's use the formula for determining the amount of progression:
s(n) = (2∙a1 + d∙(n-1))∙n/2
First, we determine the sum of the values of 101 terms of the progression by substituting the given conditions of our problem into the formula:
s 101 = (2∙0 + 0.5∙(101-1))∙101/2 = 2,525
Obviously, in order to find out the sum of the terms of the progression from the 56th to the 101st, it is necessary to subtract S 55 from S 101.
s 55 = (2∙0 + 0.5∙(55-1))∙55/2 = 742.5
Thus, the sum of the arithmetic progression for this example is:
s 101 - s 55 = 2,525 - 742.5 = 1,782.5
Example of practical application of arithmetic progression
At the end of the article, let's return to the example of an arithmetic sequence given in the first paragraph - a taximeter (taxi car meter). Let's consider this example.
Boarding a taxi (which includes 3 km of travel) costs 50 rubles. Each subsequent kilometer is paid at the rate of 22 rubles/km. Travel distance is 30 km. Calculate the cost of the trip.
1. Let’s discard the first 3 km, the price of which is included in the cost of landing.
30 - 3 = 27 km.
2. Further calculation is nothing more than parsing an arithmetic number series.
Member number - the number of kilometers traveled (minus the first three).
The value of the member is the sum.
The first term in this problem will be equal to a 1 = 50 rubles.
Progression difference d = 22 r.
the number we are interested in is the value of the (27+1)th term of the arithmetic progression - the meter reading at the end of the 27th kilometer is 27.999... = 28 km.
a 28 = 50 + 22 ∙ (28 - 1) = 644
Calendar data calculations for an arbitrarily long period are based on formulas describing certain numerical sequences. In astronomy, the length of the orbit is geometrically dependent on the distance of the celestial body to the star. In addition, various number series are successfully used in statistics and other applied areas of mathematics.
Another type of number sequence is geometric
Geometric progression is characterized by greater rates of change compared to arithmetic progression. It is no coincidence that in politics, sociology, and medicine, in order to show the high speed of spread of a particular phenomenon, for example, a disease during an epidemic, they say that the process develops in geometric progression.
The Nth term of the geometric number series differs from the previous one in that it is multiplied by some constant number - the denominator, for example, the first term is 1, the denominator is correspondingly equal to 2, then:
n=1: 1 ∙ 2 = 2
n=2: 2 ∙ 2 = 4
n=3: 4 ∙ 2 = 8
n=4: 8 ∙ 2 = 16
n=5: 16 ∙ 2 = 32,
b n - the value of the current term of the geometric progression;
b n+1 - formula of the next term of the geometric progression;
q is the denominator of the geometric progression (a constant number).
If the graph of an arithmetic progression is a straight line, then a geometric progression paints a slightly different picture:
As in the case of arithmetic, geometric progression has a formula for the value of an arbitrary term. Any nth term of a geometric progression is equal to the product of the first term and the denominator of the progression to the power of n reduced by one:
Example. We have a geometric progression with the first term equal to 3 and the denominator of the progression equal to 1.5. Let's find the 5th term of the progression
b 5 = b 1 ∙ q (5-1) = 3 ∙ 1.5 4 = 15.1875
The sum of a given number of terms is also calculated using a special formula. The sum of the first n terms of a geometric progression is equal to the difference between the product of the nth term of the progression and its denominator and the first term of the progression, divided by the denominator reduced by one:
If b n is replaced using the formula discussed above, the value of the sum of the first n terms of the number series under consideration will take the form:
Example. The geometric progression starts with the first term equal to 1. The denominator is set to 3. Let's find the sum of the first eight terms.
s8 = 1 ∙ (3 8 -1) / (3-1) = 3 280
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:
- 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 at the same time a sign of progression, therefore 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.
