Task 1 You can rent a boat at the tourist center. The rental cost is determined as follows: for the first hour you need to pay 100 rubles, and for each subsequent hour (full or incomplete) - 55 rubles. How many rubles should you pay for a boat rented for one hour, two hours, three hours, etc.?
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Problem 1* You can rent a boat at the tourist center. The rental cost is determined as follows: for the first hour you need to pay 100 rubles, and for each subsequent hour (full or incomplete) - 55 rubles. How many rubles should I pay for a boat rented for two days?
What the main point formulas?
This formula allows you to find any BY HIS NUMBER " n" .
Of course, you also need to know the first term a 1 and progression difference d, well, without these parameters you can’t write down a specific progression.
Memorizing (or cribing) this formula is not enough. You need to understand its essence and apply the formula in various problems. And also not to forget at the right moment, yes...) How not forget- I don't know. And here how to remember If necessary, I will definitely advise you. For those who complete the lesson to the end.)
So, let's look at the formula for the nth term of an arithmetic progression.
What is a formula in general? By the way, take a look if you haven’t read it. Everything is simple there. It remains to figure out what it is nth term.
Progression in general view 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- the 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 - s a 120.
How can we define it in general terms? any term of an arithmetic progression, with any number? Very simple! Like this:
a n
That's what it is nth term of an arithmetic progression. The letter n hides all the member numbers 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 progression. Using the notation a n, we can quickly find any member any arithmetic progression. And solve a bunch of other progression problems. You'll see for yourself further.
In the formula for the nth term of an arithmetic progression:
| a n = a 1 + (n-1)d |
a 1- the first term of an arithmetic progression;
n- member number.
The formula connects the key parameters of any progression: a n ; a 1 ; d And n. All progression problems revolve around these parameters.
The nth term formula can also be used to write a specific progression. For example, the problem may say that the progression is specified by the condition:
a n = 5 + (n-1) 2.
Such a problem can be a dead end... There is neither a series nor a difference... But, comparing the condition with the formula, it is easy to understand that in this progression a 1 =5, and d=2.
And it can be even worse!) If we take the same condition: a n = 5 + (n-1) 2, Yes, open the parentheses and bring similar ones? We get a new formula:
a n = 3 + 2n.
This Just not general, but for a specific progression. This is where the pitfall lurks. Some people think that the first term is a three. Although in reality the first term is five... A little lower we will work with such a modified formula.
In progression problems there is another notation - a n+1. This is, as you guessed, the “n plus first” term of the progression. Its meaning is simple and harmless.) This is a member of the progression whose number is greater than number n by one. For example, if in some problem we take a n fifth term then a n+1 will be the sixth member. Etc.
Most often the designation a n+1 found in recurrence formulas. Don't be afraid of this scary word!) This is just a way of expressing a member of an arithmetic progression through the previous one. Let's say we are given an arithmetic progression in this form, using a 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. How can we immediately count, say, the twentieth term? a 20? But there’s no way!) Until we find out the 19th term, we can’t count the 20th. This is it fundamental difference recurrent formula from the formula of the nth term. Recurrent works only through previous term, and the formula of the nth term is through first and allows straightaway find any member by its number. Without calculating the entire series of numbers in order.
In an arithmetic progression, it is easy to turn a recurrent formula 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 its usual form, and work with it. Such tasks are often encountered in the State Academy of Sciences.
Application of the formula for the nth term 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:
An arithmetic progression (a n) is given. 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 an arithmetic progression. Add and add... An hour or two.)
And according to the formula, the solution will take less than a minute. You can time it.) Let's decide.
The conditions provide all the data for using the formula: a 1 =3, d=1/6. It remains to figure out what is equal n. No problem! We need to find a 121. So 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 ours n. This is the meaning n= 121 we will substitute further into the formula, in brackets. We substitute all the numbers into the formula and calculate:
a 121 = 3 + (121-1) 1/6 = 3+20 = 23
That's it. Just as quickly one could find the five hundred and tenth term, and the thousand and third, any one. We put instead n desired number in the index of the letter " a" and in brackets, and we count.
Let me remind you the point: this formula allows you to find any arithmetic progression term BY HIS NUMBER " n" .
Let's solve the problem in a more cunning way. Let us come across 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 tell you the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Write down with your hands, 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... Is that it? If you think that’s it, then you won’t solve the problem, yes...
We still have a number n! In condition a 17 =-2 hidden two parameters. This is both the value of the seventeenth term (-2) and its number (17). Those. n=17. This “trifle” often slips past the head, and without it, (without the “trifle”, not the head!) the problem cannot be solved. Although... and without a head too.)
Now we can simply stupidly substitute our data into the formula:
a 17 = a 1 + (17-1)·(-0.5)
Oh yes, a 17 we know it's -2. Okay, let's substitute:
-2 = a 1 + (17-1)·(-0.5)
That's basically all. It remains to express the first term of the arithmetic progression from the formula and calculate it. The answer will be: a 1 = 6.
This technique - writing down a formula and simply substituting known data - is a great help in simple tasks. Well, of course, you must be able to express a variable from a formula, but what to do!? Without this skill, mathematics may not be studied at all...
Another popular puzzle:
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 are writing the formula!)
| a n = a 1 + (n-1)d |
Let's consider what we know: a 1 =2; a 15 =12; and (I’ll especially highlight!) n=15. Feel free to substitute this into the formula:
12=2 + (15-1)d
We do the arithmetic.)
12=2 + 14d
d=10/14 = 5/7
This is the correct answer.
So, the tasks for a n, a 1 And d decided. All that remains is to learn how to find the number:
The number 99 is a member of the arithmetic progression (a n), where a 1 =12; d=3. Find this member's number.
We substitute the quantities known to us 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- this is some member of the progression with a number n...And we know this member of the progression! It's 99. We don't know its number. n, So this number is what you need to find. We substitute the term of the progression 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 whether the number 117 is a member of the arithmetic progression (a n):
-3,6; -2,4; -1,2 ...
Let's write the formula again. What, there are no parameters? Hm... Why are we given eyes?) Do we see the first term of the progression? We see. This is -3.6. You can safely write: a 1 = -3.6. Difference d Can you tell from the series? It’s easy if you know what the difference of an arithmetic progression is:
d = -2.4 - (-3.6) = 1.2
So, we did the simplest thing. It remains to deal with the unknown number n and the 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... What to do!? Well, what to do, what to do... Turn on Creative skills!)
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 can we draw? Yes! Number 117 is not member of our progression. It is somewhere between the one hundred and first and one hundred and second terms. If the number turned out natural, i.e. is a positive integer, then the number would be a member of the progression with the number found. And in our case, the answer to the problem will be: No.
Task based real option GIA:
An arithmetic progression is given by the condition:
a n = -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 for the nth term 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 the arithmetic progression in it hidden. It’s okay, we’ll find it now.)
Just as in previous problems, we substitute n=1 into this formula:
a 1 = -4 + 6.8 1 = 2.8
Here! The first term is 2.8, not -4!
We look for the tenth term in the same way:
a 10 = -4 + 6.8 10 = 64
That's it.
And now, for those who have read to these lines, the promised bonus.)
Suppose, in a difficult combat situation at the State Examination or Unified State Examination, you have forgotten the useful formula for the nth term of an arithmetic progression. I remember something, but somehow uncertainly... Or n there, or n+1, or n-1... How to be!?
Calm! This formula is easy to derive. Not very strictly, but for confidence and the right decision definitely enough!) To make a conclusion, it is enough to remember the elementary meaning of an arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.
Draw a number line and mark the first one on it. second, third, etc. members. And we note the difference d between members. Like this:
We look at the picture and think: what does the second term equal? 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? It’s not for nothing that I highlight some words in bold. 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, to the number n, number of spaces will n-1. Therefore, the formula will be (without variations!):
| 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 insert a picture into the equation...
Tasks for independent solution.
To warm up:
1. In arithmetic progression (a n) a 2 =3; a 5 =5.1. Find a 3 .
Hint: according to the picture, the problem can be solved in 20 seconds... According to the formula, it turns out more difficult. But for mastering the formula, it’s more useful.) In Section 555, this problem is solved using both the picture and the formula. Feel the difference!)
And this is no longer a warm-up.)
2. In arithmetic progression (a n) a 85 =19.1; a 236 =49, 3. Find a 3 .
What, you don’t want to draw a picture?) Of course! Better according to the formula, yes...
3. The arithmetic progression is given by the condition:a 1 = -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 specified in a recurrent manner. But counting to the one hundred and twenty-fifth term... Not everyone can do such a feat.) But the formula for 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 conditions of task 4, find the sum of the smallest positive and largest negative terms of the progression.
6. The product of the fifth and twelfth terms of an increasing arithmetic progression is equal to -2.5, and the sum of the third and eleventh terms is equal to zero. Find a 14 .
Not the easiest task, yes...) The “fingertip” method won’t work here. You will 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? Happens. By the way, there is one subtle point in the last task. Care will be required when reading the problem. And logic.
The solution to all these problems is discussed in detail in Section 555. And the element of fantasy for the fourth, and the subtle point for the sixth, and general approaches for solving any problems involving the formula of the nth term - everything is described. 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. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
Lesson plan on the topic: “Formulanth member geometric progression" Preparation for the OGE.
primary goal: consolidate the concept of geometric progression;
introduce students to the formula for the nth term of a geometric progression;
application of this formula and properties using examples and problems.
UMK: Algebra.9th grade. Textbook for students of general education institutions/ (A.G. Mordkovich and others); edited by A.G. Mordkovich. - 11th ed., St.-M.: Mnemozina, 2009. - 255 pp.: ill.
Class: 9
Lesson type: lesson of learning new material.
During the classes.
Organizing time (1 min)
The teacher greets the children.
Oral work (9 min)
Find the geometric mean of the numbers 16 and 25; 9 and 36; 49 and 81; 12 and 25.
Solve the equation: b 2 =3, b 2 =-3, b 3 =-27, x 6 =164.
There is a radioactive substance weighing 256 g, the weight of which is halved per day. What will be the mass of the substance on the second day? On the third day? On the eighth day? (256; 128; 64; 32; 16; 8; 4; 2; 1;…)
You and I see that the sequence we obtained is... a geometric progression. Let's remember its definition.
A definition is given : Geometric progression is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number.
Question: - How is the second term of the sequence obtained? Third? Eighth? (By dividing the previous term by 2 or multiplying by12 ). This number is called denominator of the geometric progressions and denote q .
Examination homework(5 minutes)
Learning new material.(10 min)
Write down the sequence that corresponds to the conditions of the problem.
IN favorable conditions The bacteria multiply so that within one minute each of them divides into two. How many bacteria appeared at the 5th minute? (see Fig. 1)
How many will there be in three minutes?
At the 1st minute - 2
at the 2nd minute - 4
at the 3rd minute - 8
at the 4th minute - 16
at the 5th minute - 32
Can we continue?
at the 6th minute - 64
at the 7th minute - 128
at the 8th minute - 256
at the 9th minute - 512
at the 10th minute - 1024
at the 11th minute - 2048
at the 12th minute - 4096
at the 13th minute - 8192
Conclusion: therefore, a formula is needed to find the nth term of the geometric progression.
Consider the geometric progression b 1 ; b 2 ; b 3,...,b n, with denominator q. We have:
b 1 = b 1
b 3 = b 2 q = (b 1 q) q = b 1 q 2
b 4 = b 3 q = (b 1 q 2) q = b 1 q 3
b 5 = b 3 q = (b 1 q 3) q = b 1 q 4 etc.
It is not difficult to guess that for any n the following inequality is true:
b n = b 1 q n - 1
Thisnth term of a geometric progression.
Let's try to check the validity of this formula for the problem with bacteria already known to us. Let's count the 5th term of the sequence
b n = b 1 q n - 1= b 5 = b 1 q 5-1 = 1·2 4 = 1·16=16.
b n = b 1 q n - 1 = b 11 = b 1 q 11-1 = 1·2 10 = 1·1024=1024.
Reinforcement of learned material: (10)
PR Example 1-2.
UC: No. 17.10(a,b),
No. 17.11(a,b),
No. 17.12(a,b)
Physical education minute (1 min)
Preparation for the OGE. (15 minutes)
Cards
Homework: (1 min.)
No. 17.10(v,d), 17.12(v,d), 17.14, 17.16
Lesson summary (1 min)
Task No.1
To find the amount arithmetic progression we have two formulas.
progression difference.
d=a2-a1=-5-(-7)=2.
Let's put everything into the formula:
S50=50*(2*(-7)+(50-1)*2)/2=50*(-14+98)/2=50*42=2100
Answer: S50=2100
Task No.2
d=a2-a1=3-1=2.
Let's put everything into the formula:
S60=60*(2*1+(60-1)*2)/2=30*(2+118)=30*120=3600
Answer: S60=3600
Task No.3
Knowing that an+1=an+4, i.e. a10=a9+4, you can, of course, calculate all the first 10 terms of the sequence, but this is labor-intensive. Moreover, if it were necessary to calculate the 300th term, it would take a very long time.
There is an easier way:
IN arithmetic progression an=a1+(n-1)d, only d is unknown to us. It can be calculated using the formula: d=an+1-an
Using this formula and the problem condition, we see that d=4. Then:
a10=a1+(10-1)4
a10=3+9*4=39.Answer: a10=39
Task No.4
Knowing that bn+1=1/2*bn, i.e. b7=1/2*b6, you can, of course, calculate all the first 7 terms of the sequence, but this is labor-intensive. Moreover, if it were necessary to calculate the 300th term, it would take a very long time.
There is an easier way:
IN geometric progression bn=b1qn-1, only q is unknown to us. It can be calculated using the formula: bn+1/bn=q
Using this formula and the problem condition, we see that q=1/2. Then:
b7=b1(1/2)(7-1)
b7=-128*(1/2)6=-128*1/64=-2.
Answer: b7=-2
Task No.5
To find the sum of the first 4 terms of a given geometric progression, let's use formulas. In our case, it is more convenient to use the first one. To do this, you need to find out b1 - the first term of the progression and q - progression denominator.
b1=62.5*21=125 (from the problem conditions). And q=2.
Then S4=125*(1-24)/(1-2)=125*(1-16)/(-1)=125*15=1875
Answer: S4=1875
Task No. 6
In a geometric progression, the sum of the first and second terms is 75, and the sum of the second and third terms is 150. Find the first three terms of this progression.
bn=b1qn-1
Then b2=b1q2-1=b1q
By condition:
1) b1+b2=75
b1+b1q=75
b1(1+q)=75
2) b2+b3=150
b1q+b1q2=150
b1(q+q2)=150
b1(q+1)q=150
We substitute from point 1)
75q=150 = q=2, then b1(1+2)=75 = b1=25
b2=25*2=50
b3=25*22=100
Answer: b1=25, b2=50, b3=100
Task No.7
In this case, instead of using formulas For geometric progression, it is easier to solve this problem head-on. Those. find b2, b3, ..., b7.
b1=64 (by condition).
b2=b1*1/2=64*1/2=64/2=32
b3=b2*1/2=32/2=16
b4=16/2=8
b5=8/2=4
b6=4/2=2
b7=2/2=1 Answer: b7=1
| Card 1 №1 . Given an arithmetic progression: -7; -5; -3; ... Find the sum of its first fifty terms. №2 . Given an arithmetic progression: 1; 3; 5; … . Find the sum of its first sixty terms. №3. The arithmetic progression (a n) is given by the conditions: a 1 =3, a n+1 =a n +4. Find a 10. №4. Geometric progression (b n) is specified by the conditions: b 1 = –128, b n+1 =1/2*b n. Find b 7 . №5. The geometric progression is specified by the condition b n =62.5*2 n. Find the sum of its first 4 terms. №6 №7. Geometric progression (b n) is specified by the conditions: b 1 =64, b n+1 =b n *1/2. Find b 7 . |
| Card 1 №1 . Given an arithmetic progression: -7; -5; -3; ... Find the sum of its first fifty terms. №2 . Given an arithmetic progression: 1; 3; 5; … . Find the sum of its first sixty terms. №3. The arithmetic progression (a n) is given by the conditions: a 1 =3, a n+1 =a n +4. Find a 10. №4. Geometric progression (b n) is specified by the conditions: b 1 = –128, b n+1 =1/2*b n. Find b 7 . №5. The geometric progression is specified by the condition b n =62.5*2 n. Find the sum of its first 4 terms. №6 . In a geometric progression, the sum of the first and second terms is 75, and the sum of the second and third terms is 150. Find the first three terms of this progression. № 7. Geometric progression (bn) is given by the conditions: b 1 =64,bn+1=bn*1/2. Find b 7 . |
Problem No. 3 of 127. Problem number on WWW.FIPI.RU - 1C5D03
Show solution to problem
Given an arithmetic progression: -6; -2; 2; ... Find the sum of its first fifty terms.
To find the amount arithmetic progression we have two formulas.
We don’t know a50, so we’ll use the second formula. To do this, we find d - progression difference.
d=a2-a1=-2-(-6)=4.
Let's put everything into the formula:
S50=50*(2*(-6)+(50-1)*4)/2=50*(-12+196)/2=50*92=4600
Answer: S50=4600
Problem No. 4 of 127. Problem number on WWW.FIPI.RU - FD1ABB
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Given an arithmetic progression: -1; 2; 5; … . Find the sum of its first fifty-five terms.
To find the amount arithmetic progression we have two formulas.
We don’t know a55, so we’ll use the second formula. To do this, we find d - progression difference.
d=a2-a1=2-(-1)=3.
Let's put everything into the formula:
S55=55*(2*(-1)+(55-1)*3)/2=55*(-2+162)/2=55*80=4400
Answer: S55=4400
Problem No. 19 of 127. Problem number on WWW.FIPI.RU - 34D7F8
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The first three terms of the arithmetic progression are written out: 20; 17; 14. What number is in 91st place in this arithmetic progression?
nth term arithmetic progression equals a1+(n-1)d
a1=20
d=a2-a1=17-20=-3
a91=a1+(n-1)d=20+(91-1)(-3)=20-270=-250
Answer: a91=-250
Problem No. 22 of 127. Problem number on WWW.FIPI.RU - 4CBA5B
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The first three terms of the arithmetic progression are written: -4; 2; 8; ... What number is in 81st place in this arithmetic progression?
nth term arithmetic progression equals a1+(n-1)d
a1=-4
d=a2-a1=2-(-4)=6
a81=a1+(n-1)d=-4+(81-1)6=-4+480=476
Answer: a81=476
Problem No. 79 of 127. Problem number on WWW.FIPI.RU - 4C12DC
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The first few terms of the arithmetic progression are written out: -7; -5; -3; ...Find her sixteenth penis.
nth term arithmetic progression equals a1+(n-1)d
a1=-7 (by condition)
a2=-5 (by condition)
d=a2-a1=-5-(-7)=2
a16=a1+(n-1)d=-7+(16-1)2=-7+30=23
Answer: a16=23
Problem No. 82 of 127. Problem number on WWW.FIPI.RU - 4D6C7C
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Given a geometric progression (b n), the denominator of which is 2, b 1 =16. Find b 4 .
Every member geometric progression can be expressed through the first term.
bn=b1qn-1
Therefore, b4=b1q4-1=b1q3=16*23=16*8=128
Answer: 128
The time deposit placed in the savings bank increased annually by 5%. What will the contribution be after 8 years, if at the beginning it was equal to 1000 rubles? (1000; 1050; 1102.5; 1157.625;…) Question: How is the second term of the sequence obtained? Third? Eighth? (Multiplying the previous one by 1.05).