The concept of a number sequence implies that each natural number corresponds to some real value. Such a series of numbers can be either arbitrary or have certain properties - a progression. In the latter case, each subsequent element (member) of the sequence can be calculated using the previous one.
Arithmetic progression– a sequence of numerical values in which its neighboring members differ from each other by same number(all elements of the series, starting from the 2nd, have a similar property). This number - the difference between the previous and subsequent terms - is constant and is called the progression difference.
Progression difference: definition
Consider a sequence consisting of j values A = a(1), a(2), a(3), a(4) ... a(j), j belongs to the set natural numbers N. Arithmetic progression, according to its definition, is a sequence in which a(3) – a(2) = a(4) – a(3) = a(5) – a(4) = … = a(j) – a(j-1) = d. The value d is the desired difference of this progression.
d = a(j) – a(j-1).
Highlight:
- An increasing progression, in which case d > 0. Example: 4, 8, 12, 16, 20, ...
- Decreasing progression, then d< 0. Пример: 18, 13, 8, 3, -2, …
Difference progression and its arbitrary elements
If 2 arbitrary terms of the progression are known (i-th, k-th), then the difference for a given sequence can be determined based on the relationship:
a(i) = a(k) + (i – k)*d, which means d = (a(i) – a(k))/(i-k).
Difference of progression and its first term
This expression will help determine an unknown value only in cases where the number of the sequence element is known.
Progression difference and its sum
The sum of a progression is the sum of its terms. To calculate the total value of its first j elements, use the appropriate formula:
S(j) =((a(1) + a(j))/2)*j, but since a(j) = a(1) + d(j – 1), then S(j) = ((a(1) + a(1) + d(j – 1))/2)*j=(( 2a(1) + d(– 1))/2)*j.
Before we start deciding arithmetic progression problems, let's look at what it is number sequence, since arithmetic progression is special case number sequence.
A number sequence is a number set, each element of which has its own serial number . The elements of this set are called members of the sequence. The serial number of a sequence element is indicated by an index:
The first element of the sequence;
The fifth element of the sequence;
- the “nth” element of the sequence, i.e. element "standing in queue" at number n.
There is a relationship between the value of a sequence element and its sequence number. Therefore, we can consider a sequence as a function whose argument is the ordinal number of the element of the sequence. In other words, we can say that the sequence is a function of the natural argument:
The sequence can be set in three ways:
1 . The sequence can be specified using a table. In this case, we simply set the value of each member of the sequence.
For example, Someone decided to take up personal time management, and to begin with, count how much time he spends on VKontakte during the week. By recording the time in the table, he will receive a sequence consisting of seven elements:
The first line of the table indicates the number of the day of the week, the second - the time in minutes. We see that, that is, on Monday Someone spent 125 minutes on VKontakte, that is, on Thursday - 248 minutes, and, that is, on Friday only 15.
2 . The sequence can be specified using the nth term formula.
In this case, the dependence of the value of a sequence element on its number is expressed directly in the form of a formula.
For example, if , then
To find the value of a sequence element with a given number, we substitute the element number into the formula of the nth term.
We do the same thing if we need to find the value of a function if the value of the argument is known. We substitute the value of the argument into the function equation:
If, for example, 
, That
Let me note once again that in a sequence, unlike an arbitrary numerical function, the argument can only be a natural number.
3 . The sequence can be specified using a formula that expresses the dependence of the value of the sequence member number n on the values of the previous members. In this case, it is not enough for us to know only the number of the sequence member to find its value. We need to specify the first member or first few members of the sequence.
For example, consider the sequence 
, 
We can find the values of sequence members in sequence, starting from the third:
That is, every time, to find the value of the nth term of the sequence, we return to the previous two. This method of specifying a sequence is called recurrent, from the Latin word recurro- come back.
Now we can define an arithmetic progression. An arithmetic progression is a simple special case of a number sequence.
Arithmetic progression is a numerical sequence, each member of which, starting from the second, is equal to the previous one added to the same number.
The number is called difference of arithmetic progression. The difference of an arithmetic progression can be positive, negative, or equal to zero.
If title="d>0">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} increasing.
For example, 2; 5; 8; eleven;...
If , then each term of an arithmetic progression is less than the previous one, and the progression is decreasing.
For example, 2; -1; -4; -7;...
If , then all terms of the progression are equal to the same number, and the progression is stationary.
For example, 2;2;2;2;...
The main property of an arithmetic progression:
Let's look at the 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 
Arithmetic progression name a sequence of numbers (terms of a progression)
In which each subsequent term differs from the previous one by a new term, which is also called step or progression difference.
Thus, by specifying the progression step and its first term, you can find any of its elements using the formula
Properties of an arithmetic progression
1) Each member of an arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next members of the progression
The converse is also true. If the arithmetic mean of adjacent odd (even) terms of a progression is equal to the term that stands between them, then this sequence of numbers is an arithmetic progression. Using this statement, it is very easy to check any sequence.
Also, by the property of arithmetic progression, the above formula can be generalized to the following
This is easy to verify if you write the terms to the right of the equal sign
It is often used in practice to simplify calculations in problems.
2) The sum of the first n terms of an arithmetic progression is calculated using the formula
Remember well the formula for the sum of an arithmetic progression; it is indispensable in calculations and is quite often found in simple life situations.
3) If you need to find not the whole sum, but part of the sequence starting from its kth term, then the following sum formula will be useful to you
4) Of practical interest is finding the sum of n terms of an arithmetic progression starting from the kth number. To do this, use the formula
On this theoretical material ends and we move on to solving common problems in practice.
Example 1. Find the fortieth term of the arithmetic progression 4;7;...
Solution:
According to the condition we have
Let's determine the progression step
Using a well-known formula, we find the fortieth term of the progression
Example 2. An arithmetic progression is given by its third and seventh terms. Find the first term of the progression and the sum of ten.
Solution:
Let us write down the given elements of the progression using the formulas
We subtract the first from the second equation, as a result we find the progression step
We substitute the found value into any of the equations to find the first term of the arithmetic progression
We calculate the sum of the first ten terms of the progression
Without using complex calculations, we found all the required quantities.
Example 3. An arithmetic progression is given by the denominator and one of its terms. Find the first term of the progression, the sum of its 50 terms starting from 50 and the sum of the first 100.
Solution:
Let's write down the formula for the hundredth element of the progression
and find the first one
Based on the first, we find the 50th term of the progression
Finding the sum of the part of the progression
and the sum of the first 100
The progression amount is 250.
Example 4.
Find the number of terms of an arithmetic progression if:
a3-a1=8, a2+a4=14, Sn=111.
Solution:
Let's write the equations in terms of the first term and the progression step and determine them
We substitute the obtained values into the sum formula to determine the number of terms in the sum
We carry out simplifications
and solve the quadratic equation
Of the two values found, only the number 8 fits the problem conditions. Thus, the sum of the first eight terms of the progression is 111.
Example 5.
Solve the equation
1+3+5+...+x=307.
Solution: This equation is the sum of an arithmetic progression. Let's write out its first term and find the difference in progression
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 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:
|
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: \(d=49-62=-13\). |
|
|
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:
|
|
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:
|
We need to find the sum of the first six terms of the progression. But we do not 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: \(n=1\); \(a_(1+1)=a_1+5=-11+5=-6\) |
|
|
\(S_6=a_1+a_2+a_3+a_4+a_5+a_6=\) |
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:
|
\(S_(25)=\)\(\frac(a_1+a_(25))(2 )\) \(\cdot 25\) |
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 everything necessary information for solving 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’d like to substitute \(d\) into the formula for the sum... and here it comes up small nuance– we don’t 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 Above 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, we add the four 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.
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, the fractional number can be entered in the form of a decimal fraction (\(2.5\)) and in the form 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.
This way you can conduct your own training and/or training of yours. younger brothers or sisters, while the level of education in the field of problems being solved 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 numerical fraction The numerator is separated from the denominator by a division sign: /
Input:
Result: \(-\frac(2)(3)\)
The whole part is separated from the fraction by the ampersand sign: &
Input:
Result: \(-1\frac(2)(3)\)
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A little theory.
Number sequence
IN everyday practice numbering is often used various items to indicate the order in which they appear. 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 the sequence of squares of 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. More exact value is equal to \(365\frac(1)(4)\) 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:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)