Before we start to decide arithmetic progression problems, consider what a number sequence is, since an arithmetic progression is a special case of a number sequence.
A numerical sequence is a numerical set, each element of which has its own serial number. The elements of this set are called members of the sequence. The ordinal number of a sequence element is indicated by an index:
The first element of the sequence;
The fifth element of the sequence;
- "nth" element of the sequence, i.e. the element "standing in the queue" at number n.
There is a dependency between the value of a sequence element and its ordinal number. Therefore, we can consider a sequence as a function whose argument is the ordinal number of an element of the sequence. In other words, one can say that the sequence is a function of the natural argument:
The sequence can be specified 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 do personal time management, and to begin with, to calculate how much time he spends on VKontakte during the week. By writing the time in a table, he will get a sequence consisting of seven elements:
The first line of the table contains 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 member formula.
In this case, the dependence of the value of a sequence element on its number is expressed directly as 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 for the nth member.
We do the same 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 instead in the equation of the function:
If, for example, 
, then
Once again, I note that in a sequence, in contrast to an arbitrary numeric function, only a natural number can be an argument.
3 . The sequence can be specified using a formula that expresses the dependence of the value of the member of the sequence with number n on the value of the previous members. In this case, it is not enough for us to know only the number of a sequence member in order 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 the members of a sequence in sequence, starting from the third:
That is, each time to find the value of the nth member of the sequence, we return to the previous two. This way of sequencing 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 numerical sequence.
Arithmetic progression is called a numerical sequence, each member of which, starting from the second, is equal to the previous one, added with the same number.
The number is called the difference of an arithmetic progression. The difference of an arithmetic progression can be positive, negative, or zero.
If title="(!LANG:d>0">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} increasing.
For example, 2; 5; eight; eleven;...
If , then each term of the arithmetic progression is less than the previous one, and the progression is waning.
For example, 2; -one; -four; -7;...
If , then all members 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:
.
Divide both sides of the equation by 2:
So, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of two neighboring ones:
Moreover, because
, and at the same time
, then
, and hence
Each member of the arithmetic progression starting with title="(!LANG:k>l">, равен среднему арифметическому двух равноотстоящих.
!}
th member formula.
We see that for the members of the arithmetic progression, the following relations hold:
and finally
We got formula of the nth term.
IMPORTANT! Any member of an arithmetic progression can be expressed in terms of and . Knowing the first term and the difference of an arithmetic progression, you can find any of its members.
The sum of n members of an arithmetic progression.
In an arbitrary arithmetic progression, the sums of terms equally spaced from the extreme ones are equal to each other:
Consider an arithmetic progression with n members. Let the sum of n members of this progression be equal to .
Arrange the terms of the progression first in ascending order of numbers, and then in descending order:
Let's pair it up:
The sum in each parenthesis is , the number of pairs is n.
We get:
So, the sum of n members of an arithmetic progression can be found using the formulas:
Consider solving arithmetic progression problems.
1 . The sequence is given by the formula of the nth member: . Prove that this sequence is an arithmetic progression.
Let us prove that the difference between two adjacent members of the sequence is equal to the same number.
We have obtained that the difference of 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 the 31 terms of the progression.
b) Determine if 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 
If every natural number n match a real number a n , then they say that given number sequence :
a 1 , a 2 , a 3 , . . . , a n , . . . .
So, a numerical sequence is a function of a natural argument.
Number a 1 called the first member of the sequence , number a 2 — the second member of the sequence , number a 3 — third and so on. Number a n called nth member of the sequence , and the natural number n — his number .
From two neighboring members a n and a n +1 member sequences a n +1 called subsequent (towards a n ), a a n — previous (towards a n +1 ).
To specify a sequence, you must specify a method that allows you to find a sequence member with any number.
Often the sequence is given with nth term formulas , that is, a formula that allows you to determine a sequence member by its number.
For example,
the sequence of positive odd numbers can be given by the formula
a n= 2n- 1,
and the sequence of alternating 1 and -1 - formula
b n = (-1)n +1 . ◄
The sequence can be determined recurrent formula, that is, a formula that expresses any member of the sequence, starting with some, through the previous (one or more) members.
For example,
if a 1 = 1 , a a n +1 = a n + 5
a 1 = 1,
a 2 = a 1 + 5 = 1 + 5 = 6,
a 3 = a 2 + 5 = 6 + 5 = 11,
a 4 = a 3 + 5 = 11 + 5 = 16,
a 5 = a 4 + 5 = 16 + 5 = 21.
If a a 1= 1, a 2 = 1, a n +2 = a n + a n +1 , then the first seven members of the numerical sequence are set as follows:
a 1 = 1,
a 2 = 1,
a 3 = a 1 + a 2 = 1 + 1 = 2,
a 4 = a 2 + a 3 = 1 + 2 = 3,
a 5 = a 3 + a 4 = 2 + 3 = 5,
a 6 = a 4 + a 5 = 3 + 5 = 8,
a 7 = a 5 + a 6 = 5 + 8 = 13. ◄
Sequences can be final and endless .
The sequence is called ultimate if it has a finite number of members. The sequence is called endless if it has infinitely many members.
For example,
sequence of two-digit natural numbers:
10, 11, 12, 13, . . . , 98, 99
final.
Prime number sequence:
2, 3, 5, 7, 11, 13, . . .
endless. ◄
The sequence is called increasing , if each of its members, starting from the second, is greater than the previous one.
The sequence is called waning , if each of its members, starting from the second, is less than the previous one.
For example,
2, 4, 6, 8, . . . , 2n, . . . is an ascending sequence;
1, 1 / 2 , 1 / 3 , 1 / 4 , . . . , 1 /n, . . . is a descending sequence. ◄
A sequence whose elements do not decrease with increasing number, or, conversely, do not increase, is called monotonous sequence .
Monotonic sequences, in particular, are increasing sequences and decreasing sequences.
Arithmetic progression
Arithmetic progression a sequence is called, each member of which, starting from the second, is equal to the previous one, to which the same number is added.
a 1 , a 2 , a 3 , . . . , a n, . . .
is an arithmetic progression if for any natural number n condition is met:
a n +1 = a n + d,
where d - some number.
Thus, the difference between the next and the previous members of a given arithmetic progression is always constant:
a 2 - a 1 = a 3 - a 2 = . . . = a n +1 - a n = d.
Number d called the difference of an arithmetic progression.
To set an arithmetic progression, it is enough to specify its first term and difference.
For example,
if a 1 = 3, d = 4 , then the first five terms of the sequence are found as follows:
a 1 =3,
a 2 = a 1 + d = 3 + 4 = 7,
a 3 = a 2 + d= 7 + 4 = 11,
a 4 = a 3 + d= 11 + 4 = 15,
a 5 = a 4 + d= 15 + 4 = 19. ◄
For an arithmetic progression with the first term a 1 and difference d her n
a n = a 1 + (n- 1)d.
For example,
find the thirtieth term of an arithmetic progression
1, 4, 7, 10, . . .
a 1 =1, d = 3,
a 30 = a 1 + (30 - 1)d= 1 + 29· 3 = 88. ◄
a n-1 = a 1 + (n- 2)d,
a n= a 1 + (n- 1)d,
a n +1 = a 1 + nd,
then obviously
| a n=
| a n-1 + a n+1
|
| 2
|
each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the previous and subsequent members.
numbers a, b and c are consecutive members of some arithmetic progression if and only if one of them is equal to the arithmetic mean of the other two.
For example,
a n = 2n- 7 , is an arithmetic progression.
Let's use the statement above. We have:
a n = 2n- 7,
a n-1 = 2(n- 1) - 7 = 2n- 9,
a n+1 = 2(n+ 1) - 7 = 2n- 5.
Consequently,
| a n+1 + a n-1
| =
| 2n- 5 + 2n- 9
| = 2n- 7 = a n,
|
| 2
| 2
|
◄
Note that n -th member of an arithmetic progression can be found not only through a 1 , but also any previous a k
a n = a k + (n- k)d.
For example,
for a 5 can be written
a 5 = a 1 + 4d,
a 5 = a 2 + 3d,
a 5 = a 3 + 2d,
a 5 = a 4 + d. ◄
a n = a n-k + kd,
a n = a n+k - kd,
then obviously
| a n=
| a n-k
+ a n+k
|
| 2
|
any member of an arithmetic progression, starting from the second, is equal to half the sum of the members of this arithmetic progression equally spaced from it.
In addition, for any arithmetic progression, the equality is true:
a m + a n = a k + a l,
m + n = k + l.
For example,
in arithmetic progression
1) a 10 = 28 = (25 + 31)/2 = (a 9 + a 11 )/2;
2) 28 = a 10 = a 3 + 7d= 7 + 7 3 = 7 + 21 = 28;
3) a 10= 28 = (19 + 37)/2 = (a 7 + a 13)/2;
4) a 2 + a 12 = a 5 + a 9, because
a 2 + a 12= 4 + 34 = 38,
a 5 + a 9 = 13 + 25 = 38. ◄
S n= a 1 + a 2 + a 3 + . . .+ a n,
first n members of an arithmetic progression is equal to the product of half the sum of the extreme terms by the number of terms:
From this, in particular, it follows that if it is necessary to sum the terms
a k, a k +1 , . . . , a n,
then the previous formula retains its structure:
For example,
in arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, . . .
S 10 = 1 + 4 + . . . + 28 = (1 + 28) · 10/2 = 145;
10 + 13 + 16 + 19 + 22 + 25 + 28 = S 10 - S 3 = (10 + 28 ) · (10 - 4 + 1)/2 = 133. ◄
If an arithmetic progression is given, then the quantities a 1 , a n, d, n andS n linked by two formulas:
Therefore, if the values of three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.
An arithmetic progression is a monotonic sequence. Wherein:
- if d > 0 , then it is increasing;
- if d < 0 , then it is decreasing;
- if d = 0 , then the sequence will be stationary.
Geometric progression
geometric progression a sequence is called, each term of which, starting from the second, is equal to the previous one, multiplied by the same number.
b 1 , b 2 , b 3 , . . . , b n, . . .
is a geometric progression if for any natural number n condition is met:
b n +1 = b n · q,
where q ≠ 0 - some number.
Thus, the ratio of the next term of this geometric progression to the previous one is a constant number:
b 2 / b 1 = b 3 / b 2 = . . . = b n +1 / b n = q.
Number q called denominator of a geometric progression.
To set a geometric progression, it is enough to specify its first term and denominator.
For example,
if b 1 = 1, q = -3 , then the first five terms of the sequence are found as follows:
b 1 = 1,
b 2 = b 1 · q = 1 · (-3) = -3,
b 3 = b 2 · q= -3 · (-3) = 9,
b 4 = b 3 · q= 9 · (-3) = -27,
b 5 = b 4 · q= -27 · (-3) = 81. ◄
b 1 and denominator q her n -th term can be found by the formula:
b n = b 1 · q n -1 .
For example,
find the seventh term of a geometric progression 1, 2, 4, . . .
b 1 = 1, q = 2,
b 7 = b 1 · q 6 = 1 2 6 = 64. ◄
bn-1 = b 1 · q n -2 ,
b n = b 1 · q n -1 ,
b n +1 = b 1 · q n,
then obviously
b n 2 = b n -1 · b n +1 ,
each member of the geometric progression, starting from the second, is equal to the geometric mean (proportional) of the previous and subsequent members.
Since the converse is also true, the following assertion holds:
numbers a, b and c are consecutive members of some geometric progression if and only if the square of one of them is equal to the product of the other two, that is, one of the numbers is the geometric mean of the other two.
For example,
let us prove that the sequence given by the formula b n= -3 2 n , is a geometric progression. Let's use the statement above. We have:
b n= -3 2 n,
b n -1 = -3 2 n -1 ,
b n +1 = -3 2 n +1 .
Consequently,
b n 2 = (-3 2 n) 2 = (-3 2 n -1 ) (-3 2 n +1 ) = b n -1 · b n +1 ,
which proves the required assertion. ◄
Note that n th term of a geometric progression can be found not only through b 1 , but also any previous term b k , for which it suffices to use the formula
b n = b k · q n - k.
For example,
for b 5 can be written
b 5 = b 1 · q 4 ,
b 5 = b 2 · q 3,
b 5 = b 3 · q2,
b 5 = b 4 · q. ◄
b n = b k · q n - k,
b n = b n - k · q k,
then obviously
b n 2 = b n - k· b n + k
the square of any member of a geometric progression, starting from the second, is equal to the product of the members of this progression equidistant from it.
In addition, for any geometric progression, the equality is true:
b m· b n= b k· b l,
m+ n= k+ l.
For example,
exponentially
1) b 6 2 = 32 2 = 1024 = 16 · 64 = b 5 · b 7 ;
2) 1024 = b 11 = b 6 · q 5 = 32 · 2 5 = 1024;
3) b 6 2 = 32 2 = 1024 = 8 · 128 = b 4 · b 8 ;
4) b 2 · b 7 = b 4 · b 5 , because
b 2 · b 7 = 2 · 64 = 128,
b 4 · b 5 = 8 · 16 = 128. ◄
S n= b 1 + b 2 + b 3 + . . . + b n
first n members of a geometric progression with a denominator q ≠ 0 calculated by the formula:
And when q = 1 - according to the formula
S n= n.b. 1
Note that if we need to sum the terms
b k, b k +1 , . . . , b n,
then the formula is used:
| S n- S k -1 = b k + b k +1 + . . . + b n = b k · | 1 - q n -
k +1
| . |
| 1 - q
|
For example,
exponentially 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, . . .
S 10 = 1 + 2 + . . . + 512 = 1 · (1 - 2 10) / (1 - 2) = 1023;
64 + 128 + 256 + 512 = S 10 - S 6 = 64 · (1 - 2 10-7+1) / (1 - 2) = 960. ◄
If a geometric progression is given, then the quantities b 1 , b n, q, n and S n linked by two formulas:
Therefore, if the values of any three of these quantities are given, then the corresponding values of the other two quantities are determined from these formulas combined into a system of two equations with two unknowns.
For a geometric progression with the first term b 1 and denominator q the following take place monotonicity properties :
- the progression is increasing if one of the following conditions is met:
b 1 > 0 and q> 1;
b 1 < 0 and 0 < q< 1;
- A progression is decreasing if one of the following conditions is met:
b 1 > 0 and 0 < q< 1;
b 1 < 0 and q> 1.
If a q< 0 , then the geometric progression is sign-alternating: its odd-numbered terms have the same sign as its first term, and even-numbered terms have the opposite sign. It is clear that an alternating geometric progression is not monotonic.
Product of the first n terms of a geometric progression can be calculated by the formula:
P n= b 1 · b 2 · b 3 · . . . · b n = (b 1 · b n) n / 2 .
For example,
1 · 2 · 4 · 8 · 16 · 32 · 64 · 128 = (1 · 128) 8/2 = 128 4 = 268 435 456;
3 · 6 · 12 · 24 · 48 = (3 · 48) 5/2 = (144 1/2) 5 = 12 5 = 248 832.◄
Infinitely decreasing geometric progression
Infinitely decreasing geometric progression is called an infinite geometric progression whose denominator modulus is less than 1 , that is
|q| < 1 .
Note that an infinitely decreasing geometric progression may not be a decreasing sequence. This fits the case
1 < q< 0 .
With such a denominator, the sequence is sign-alternating. For example,
1, - 1 / 2 , 1 / 4 , - 1 / 8 , . . . .
The sum of an infinitely decreasing geometric progression name the number to which the sum of the first n terms of the progression with an unlimited increase in the number n . This number is always finite and is expressed by the formula
| S= b 1 + b 2 + b 3 + . . . = | b 1
| . |
| 1 - q
|
For example,
10 + 1 + 0,1 + 0,01 + . . . = 10 / (1 - 0,1) = 11 1 / 9 ,
10 - 1 + 0,1 - 0,01 + . . . = 10 / (1 + 0,1) = 9 1 / 11 . ◄
Relationship between arithmetic and geometric progressions
Arithmetic and geometric progressions are closely related. Let's consider just two examples.
a 1 , a 2 , a 3 , . . . d , then
b a 1 , b a 2 , b a 3 , . . . b d .
For example,
1, 3, 5, . . . — arithmetic progression with difference 2 and
7 1 , 7 3 , 7 5 , . . . is a geometric progression with a denominator 7 2 . ◄
b 1 , b 2 , b 3 , . . . is a geometric progression with a denominator q , then
log a b 1, log a b 2, log a b 3, . . . — arithmetic progression with difference log aq .
For example,
2, 12, 72, . . . is a geometric progression with a denominator 6 and
lg 2, lg 12, lg 72, . . . — arithmetic progression with difference lg 6 . ◄
What is the essence of the formula?
This formula allows you to find any BY HIS NUMBER" n" .
Of course, you need to know the first term a 1 and progression difference d, well, without these parameters, you can’t write down a specific progression.
It is not enough to memorize (or cheat) this formula. It is necessary to assimilate its essence and apply the formula in various problems. Yes, and do not forget at the right time, yes ...) How not forget- I do not know. But how to remember If needed, I'll give you a hint. For those who master the lesson to the end.)
So, let's deal with the formula of the n-th member of an arithmetic progression.
What is a formula in general - we imagine.) What is an arithmetic progression, a member number, a progression difference - is clearly stated in the previous lesson. Take a look if you haven't read it. Everything is simple there. It remains to figure out what nth member.
The progression in general can be written as a series of numbers:
a 1 , a 2 , a 3 , a 4 , a 5 , .....
a 1- denotes the first term of an arithmetic progression, a 3- third member a 4- fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - from a 120.
How to define in general any member of an arithmetic progression, s any number? Very simple! Like this:
a n
That's what it is n-th member of an arithmetic progression. Under the letter n all the numbers of members are hidden at once: 1, 2, 3, 4, and so on.
And what does such a record give us? Just think, instead of a number, they wrote down a letter ...
This notation gives us a powerful tool for working with arithmetic progressions. Using the notation a n, we can quickly find any member any arithmetic progression. And a bunch of tasks to solve in progression. You will see further.
In the formula of the nth member of an arithmetic progression:
| a n = a 1 + (n-1)d |
a 1- the first member of the arithmetic progression;
n- member number.
The formula links the key parameters of any progression: a n ; a 1 ; d and n. Around these parameters, all the puzzles revolve in progression.
The nth term formula can also be used to write a specific progression. For example, in the problem it can be said that the progression is given by the condition:
a n = 5 + (n-1) 2.
Such a problem can even confuse ... There is no series, no difference ... But, comparing the condition with the formula, it is easy to figure out that in this progression a 1 \u003d 5, and d \u003d 2.
And it can be even angrier!) If we take the same condition: a n = 5 + (n-1) 2, yes, open the brackets and give similar ones? We get a new formula:
an = 3 + 2n.
it Only not general, but for a specific progression. This is where the pitfall lies. Some people think that the first term is a three. Although in reality the first member is a five ... A little lower we will work with such a modified formula.
In tasks for progression, there is another notation - a n+1. This is, you guessed it, the "n plus the first" term of the progression. Its meaning is simple and harmless.) This is a member of the progression, the number of which is greater than the number n by one. For example, if in some problem we take for a n fifth term, then a n+1 will be the sixth member. Etc.
Most often the designation a n+1 occurs in recursive formulas. Do not be afraid of this terrible word!) This is just a way of expressing a term of an arithmetic progression through the previous one. Suppose we are given an arithmetic progression in this form, using the recurrent formula:
a n+1 = a n +3
a 2 = a 1 + 3 = 5+3 = 8
a 3 = a 2 + 3 = 8+3 = 11
The fourth - through the third, the fifth - through the fourth, and so on. And how to count immediately, say the twentieth term, a 20? But no way!) While the 19th term is not known, the 20th cannot be counted. This is the fundamental difference between the recursive formula and the formula of the nth term. Recursive works only through previous term, and the formula of the nth term - through the first and allows straightaway find any member by its number. Not counting the whole series of numbers in order.
In an arithmetic progression, a recursive formula can easily be turned into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in the usual form, and work with it. In the GIA, such tasks are often found.
Application of the formula of the n-th member of an arithmetic progression.
First, let's look at the direct application of the formula. At the end of the previous lesson there was a problem:
Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.
This problem can be solved without any formulas, simply based on the meaning of the arithmetic progression. Add, yes add ... An hour or two.)
And according to the formula, the solution will take less than a minute. You can time it.) We decide.
The conditions provide all the data for using the formula: a 1 \u003d 3, d \u003d 1/6. It remains to be seen what n. No problem! We need to find a 121. Here we write:
Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in the member of the arithmetic progression number one hundred twenty one. This will be our n. It is this meaning n= 121 we will substitute further into the formula, in brackets. Substitute all the numbers in the formula and calculate:
a 121 = 3 + (121-1) 1/6 = 3+20 = 23
That's all there is to it. Just as quickly one could find the five hundred and tenth member, and the thousand and third, any. We put instead n the desired number in the index of the letter " a" and in brackets, and we consider.
Let me remind you the essence: this formula allows you to find any term of an arithmetic progression BY HIS NUMBER" n" .
Let's solve the problem smarter. Let's say we have the following problem:
Find the first term of the arithmetic progression (a n) if a 17 =-2; d=-0.5.
If you have any difficulties, I will suggest the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Hand write, right in your notebook:
| a n = a 1 + (n-1)d |
And now, looking at the letters of the formula, we understand what data we have and what is missing? Available d=-0.5, there is a seventeenth member ... Everything? If you think that's all, then you can't solve the problem, yes ...
We also have a number n! In the condition a 17 =-2 hidden two options. This is both the value of the seventeenth member (-2) and its number (17). Those. n=17. This "little thing" often slips past the head, and without it, (without the "little thing", not the head!) The problem cannot be solved. Although ... and without a head too.)
Now we can just stupidly substitute our data into the formula:
a 17 \u003d a 1 + (17-1) (-0.5)
Oh yes, a 17 we know it's -2. Okay, let's put it in:
-2 \u003d a 1 + (17-1) (-0.5)
That, in essence, is all. It remains to express the first term of the arithmetic progression from the formula, and calculate. You get the answer: a 1 = 6.
Such a technique - writing a formula and simply substituting known data - helps a lot in simple tasks. Well, you must, of course, be able to express a variable from a formula, but what to do!? Without this skill, mathematics can not be studied at all ...
Another popular problem:
Find the difference of the arithmetic progression (a n) if a 1 =2; a 15 =12.
What are we doing? You will be surprised, we write the formula!)
| a n = a 1 + (n-1)d |
Consider what we know: a 1 =2; a 15 =12; and (special highlight!) n=15. Feel free to substitute in the formula:
12=2 + (15-1)d
Let's do the arithmetic.)
12=2 + 14d
d=10/14 = 5/7
This is the correct answer.
So, tasks a n , a 1 and d decided. It remains to learn how to find the number:
The number 99 is a member of an arithmetic progression (a n), where a 1 =12; d=3. Find the number of this member.
We substitute the known quantities into the formula of the nth term:
a n = 12 + (n-1) 3
At first glance, there are two unknown quantities here: a n and n. But a n is some member of the progression with the number n... And this member of the progression we know! It's 99. We don't know his number. n, so this number also needs to be found. Substitute the progression term 99 into the formula:
99 = 12 + (n-1) 3
We express from the formula n, we think. We get the answer: n=30.
And now a problem on the same topic, but more creative):
Determine if the number 117 will be a member of an arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
Let's write the formula again. What, there are no parameters? Hm... Why do we need eyes?) Do we see the first member of the progression? We see. This is -3.6. You can safely write: a 1 \u003d -3.6. Difference d can be determined from the series? It's easy if you know what the difference of an arithmetic progression is:
d = -2.4 - (-3.6) = 1.2
Yes, we did the simplest thing. It remains to deal with an unknown number n and an incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know that ... How to be!? Well, how to be, how to be... Turn on your creative abilities!)
We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes-yes!)) and substitute our numbers:
117 = -3.6 + (n-1) 1.2
Again we express from the formulan, we count and get:
Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion do we draw? Yes! Number 117 is not member of our progression. It is somewhere between the 101st and 102nd member. If the number turned out to be natural, i.e. positive integer, then the number would be a member of the progression with the found number. And in our case, the answer to the problem will be: no.
Task based on a real version of the GIA:
The arithmetic progression is given by the condition:
a n \u003d -4 + 6.8n
Find the first and tenth terms of the progression.
Here the progression is set in an unusual way. Some kind of formula ... It happens.) However, this formula (as I wrote above) - also the formula of the n-th member of an arithmetic progression! She also allows find any member of the progression by its number.
We are looking for the first member. The one who thinks. that the first term is minus four, is fatally mistaken!) Because the formula in the problem is modified. The first term of an arithmetic progression in it hidden. Nothing, we'll find it now.)
Just as in the previous tasks, we substitute n=1 into this formula:
a 1 \u003d -4 + 6.8 1 \u003d 2.8
Here! The first term is 2.8, not -4!
Similarly, we are looking for the tenth term:
a 10 \u003d -4 + 6.8 10 \u003d 64
That's all there is to it.
And now, for those who have read up to these lines, the promised bonus.)
Suppose, in a difficult combat situation of the GIA or the Unified State Exam, you forgot the useful formula of the n-th member of an arithmetic progression. Something comes to mind, but somehow uncertainly ... Whether n there, or n+1, or n-1... How to be!?
Calm! This formula is easy to derive. Not very strict, but definitely enough for confidence and the right decision!) For the conclusion, it is enough to remember the elementary meaning of the arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.
We draw a numerical axis and mark the first one on it. second, third, etc. members. And note the difference d between members. Like this:
We look at the picture and think: what is the second term equal to? Second one d:
a 2 =a 1 + 1 d
What is the third term? Third term equals first term plus two d.
a 3 =a 1 + 2 d
Do you get it? I don't put some words in bold for nothing. Okay, one more step.)
What is the fourth term? Fourth term equals first term plus three d.
a 4 =a 1 + 3 d
It's time to realize that the number of gaps, i.e. d, always one less than the number of the member you are looking for n. That is, up to the number n, number of gaps will be n-1. So, the formula will be (no options!):
| a n = a 1 + (n-1)d |
In general, visual pictures are very helpful in solving many problems in mathematics. Don't neglect the pictures. But if it's difficult to draw a picture, then ... only a formula!) In addition, the formula of the nth term allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. You can't put a picture in an equation...
Tasks for independent decision.
For warm-up:
1. In arithmetic progression (a n) a 2 =3; a 5 \u003d 5.1. Find a 3 .
Hint: according to the picture, the problem is solved in 20 seconds ... According to the formula, it turns out more difficult. But for mastering the formula, it is more useful.) In Section 555, this problem is solved both by the picture and by the formula. Feel the difference!)
And this is no longer a warm-up.)
2. In arithmetic progression (a n) a 85 \u003d 19.1; a 236 =49, 3. Find a 3 .
What, reluctance to draw a picture?) Still! It's better in the formula, yes ...
3. Arithmetic progression is given by the condition:a 1 \u003d -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.
In this task, the progression is given in a recurrent way. But counting up to the one hundred and twenty-fifth term... Not everyone can do such a feat.) But the formula of the nth term is within the power of everyone!
4. Given an arithmetic progression (a n):
-148; -143,8; -139,6; -135,4, .....
Find the number of the smallest positive term of the progression.
5. According to the condition of task 4, find the sum of the smallest positive and largest negative members of the progression.
6. The product of the fifth and twelfth terms of an increasing arithmetic progression is -2.5, and the sum of the third and eleventh terms is zero. Find a 14 .
Not the easiest task, yes ...) Here the method "on the fingers" will not work. You have to write formulas and solve equations.
Answers (in disarray):
3,7; 3,5; 2,2; 37; 2,7; 56,5
Happened? It's nice!)
Not everything works out? It happens. By the way, in the last task there is one subtle point. Attentiveness when reading the problem will be required. And logic.
The solution to all these problems is discussed in detail in Section 555. And the fantasy element for the fourth, and the subtle moment for the sixth, and general approaches for solving any problems for the formula of the nth term - everything is painted. I recommend.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
Arithmetic and geometric progressions
Theoretical information
Theoretical information
Arithmetic progression |
Geometric progression |
|
Definition |
Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference) |
geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression) |
Recurrent formula |
For any natural n |
For any natural n |
nth term formula |
a n = a 1 + d (n - 1) |
b n \u003d b 1 ∙ q n - 1, b n ≠ 0 |
| characteristic property |  |
|
| Sum of the first n terms |  |
 |
Examples of tasks with comments
Exercise 1
In arithmetic progression ( a n) a 1 = -6, a 2
According to the formula of the nth term:
a 22 = a 1+ d (22 - 1) = a 1+ 21d
By condition:
a 1= -6, so 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 2
Find the fifth term of the geometric progression: -3; 6;....
1st way (using n-term formula)
According to the formula of the n-th member of a geometric progression:
b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.
Because b 1 = -3,
2nd way (using recursive 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:
.
Substitute the data in 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 apply in this case?
By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, and a 16 without finding d . Therefore, we 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 a geometric progression are recorded:
Find the term of the progression, denoted by the letter x .
When solving, we use the formula for the nth term b n \u003d b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.
Substituting the found values into the formula, we get:
.
Answer : .
Task 7
From the arithmetic progressions given by the formula of the nth term, choose the one for which the condition is satisfied a 27 > 9:
Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:
.
Answer: 4.
Task 8
In arithmetic progression a 1= 3, d = -1.5. Specify the largest value of n for which the inequality holds a n > -6.
Online calculator.
Arithmetic progression solution.
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, a fractional number can be entered as a decimal fraction (\(2.5 \)) and as an ordinary 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 can be useful for high school students in preparing for tests and exams, when testing knowledge before the Unified State Examination, and 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 your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.
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 dot or a comma.
For example, you can enter decimals like 2.5 or like 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: \(-\frac(2)(3) \)
The integer part is separated from the fraction by an ampersand: &
Input:
Result: \(-1\frac(2)(3) \)
It was found that some scripts needed to solve this task were not loaded, and the program may not work.
You may have AdBlock enabled.
In this case, disable it and refresh the page.
JavaScript must be enabled for the solution to appear.
Here are instructions on how to enable JavaScript in your browser.
Because There are a lot of people who want to solve the problem, your request is queued.
After a few seconds, the solution will appear below.
Wait, please sec...
If you noticed an error in the solution, then you can write about it in the Feedback Form .
Do not forget indicate which task you decide what enter in the fields.
Our games, puzzles, emulators:
A bit of theory.
Numeric sequence
In everyday practice, the numbering of various objects is often used to indicate the order in which they are located. For example, the houses on each street are numbered. In the library, reader's subscriptions are numbered and then arranged in the order of the assigned numbers in special file cabinets.
In a savings bank, by the number of the depositor's personal account, you can easily find this account and see what kind of deposit it has. Let there be a deposit of a1 rubles on account No. 1, a deposit of a2 rubles on account No. 2, etc. It turns out numerical 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 assigned a number a n .
Mathematics also studies infinite number sequences:
a 1 , a 2 , a 3 , ..., a n , ... .
The number a 1 is called the first member of the sequence, number a 2 - the second member of the sequence, number a 3 - the third member 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 member of the sequence; and n = n 2 is the nth member of the sequence; a n+1 = (n + 1) 2 is the (n + 1)th (en plus the first) member of the sequence. Often a sequence can be specified by the formula of its nth term. For example, the formula \(a_n=\frac(1)(n), \; n \in \mathbb(N) \) gives the sequence \(1, \; \frac(1)(2) , \; \frac( 1)(3) , \; \frac(1)(4) , \dots,\frac(1)(n) , \dots \)
Arithmetic progression
The length of a 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 elongated year is called a leap year.
For example, in the third millennium, leap years are 2004, 2008, 2012, 2016, ... .
In this sequence, each member, starting from the second, is equal to the previous one, added with the same number 4. Such sequences are called arithmetic progressions.
Definition.
The numerical 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.
It follows from this formula 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 member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of the two members adjacent to it. 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 recursive 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, for a 100, a lot of calculations will already be required. Usually, the nth term formula is used for this. According to the definition of an arithmetic progression
\(a_2=a_1+d, \)
\(a_3=a_2+d=a_1+2d, \)
\(a_4=a_3+d=a_1+3d\)
etc.
Generally,
\(a_n=a_1+(n-1)d, \)
since the nth member of an arithmetic progression is obtained from the first member by adding (n-1) times the number d.
This formula is called formula of the nth member of an arithmetic progression.
The sum of the first n terms of an arithmetic progression
Let's find the sum of all natural numbers from 1 to 100.
We write this sum in two ways:
S = l + 2 + 3 + ... + 99 + 100,
S = 100 + 99 + 98 + ... + 2 + 1.
We add these equalities term by term:
2S = 101 + 101 + 101 + ... + 101 + 101.
There are 100 terms in this sum.
Therefore, 2S = 101 * 100, whence S = 101 * 50 = 5050.
Consider now 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 \u003d a 1, a 2, a 3, ..., a n
Then the sum of the first n terms of an arithmetic progression is
\(S_n = n \cdot \frac(a_1+a_n)(2) \)
Since \(a_n=a_1+(n-1)d \), then replacing a n in this formula, we get another formula for finding the sums of the first n terms of an arithmetic progression:
\(S_n = n \cdot \frac(2a_1+(n-1)d)(2) \)
