Presentation for the lesson "arithmetic and geometric progression". Lesson topic: Arithmetic and geometric progression. VII. Homework
Slide 1
Arithmetic and Geometric progression
Project of 9b grade student Dmitry Tesli
Slide 2
Progression
- number sequence, each member of which, starting from the second, is equal to the previous one, added to the constant number d for this sequence. The number d is called the progression difference.
- a numerical sequence, each member of which, starting from the second, is equal to the previous one, multiplied by a constant number q for this sequence. The number q is called the denominator of the progression.
Progression
Slide 3
Arithmetic Geometric Any member arithmetic progression
is calculated by the formula: an=a1+d(n–1) The sum of the first n terms of an arithmetic progression is calculated as follows: Sn=0.5(a1+an)n Any term of a geometric progression is calculated by the formula: bn=b1qn-1 The sum of the first n members of the geometric progression is calculated as follows: Sn=b1(qn-1)/q-1
Slide 4
Arithmetic progression Known interesting story about the famous German mathematician K. Gauss (1777 - 1855), who as a child showed outstanding abilities for mathematics. The teacher asked the students to add up everything integers
from 1 to 100. Little Gauss solved this problem in one minute, realizing that the sums are 1+100, 2+99, etc. are equal, he multiplied 101 by 50, i.e. by the number of such amounts. In other words, he noticed a pattern inherent in arithmetic progressions.
Slide 5 Infinitely decreasing
geometric progression
is a geometric progression for which |q|
Slide 6
Arithmetic and geometric progressions as a justification for wars
The English economist Bishop Malthus used geometric and arithmetic progressions to justify wars: means of consumption (food, clothing) grow according to the laws of arithmetic progression, and people multiply according to the laws of geometric progression. Wars are necessary to get rid of excess population.
Slide 7
Probably the first situation in which people had to deal with geometric progression was counting the size of a herd, carried out several times at regular intervals. If no emergency occurs, the number of newborns and dead animals is proportional to the number of all animals. This means that if over a certain period of time the number of sheep a shepherd has increased from 10 to 20, then over the next same period it will double again and become equal to 40.
Slide 8
Ecology and industry
Wood growth in forests occurs according to the laws of geometric progression. Moreover, each tree species has its own coefficient of annual volume growth. Taking into account these changes makes it possible to plan the cutting down of part of the forests and simultaneous work on forest restoration.
Slide 9
Biology
A bacterium divides into three in one second. How many bacteria will be in the test tube in five seconds? The first member of the progression is one bacterium. Using the formula, we find that in the second second we will have 3 bacteria, in the third - 9, in the fourth - 27, in the fifth - 32. Thus, we can calculate the number of bacteria in the test tube at any time.
Slide 10
Economy
In life practice, geometric progression appears primarily in the problem of calculating compound interest. The time deposit placed in a savings bank increases by 5% annually. What will the contribution be after 5 years, if at first it was equal to 1000 rubles? The next year after the deposit we will have 1050 rubles, in the third year - 1102.5, in the fourth - 1157.625, in the fifth - 1215.50625 rubles.
12; 5; 8; 11;14; 17;… 2) 3; 9; 27; 81; 243;… 3) 1; 6; eleven; 20; 25;… ––––32 4) –4; -8; -16; –32; ... 5) 5; 25; 35; 45; 55;… –2–– 6– 8 6) –2; -4; – 6; - 8; ... arithmetic progression d = 3 – 2 arithmetic progression d = – 2 geometric progression q = 3 sequence of numbers geometric progression q = 2 sequence of numbers
UE2 1) Given: (a n) arithmetic progression a 1 = 5 d = 3 Find: a 6 ; a 10. Solution: using the formula a n = a 1 +(n -1) d a 6 = a 1 +5 d = = 20 a 10 = a 1 +9 d = = 32 Answer: 20; 32 Solution
UE2 1) Given: (b n) geometric progression b 1 = 5 q = 3 Find: b 3 ; b 5. Solution: using the formula b n = b 1 q n-1 b 3 =b 1 q 2 = =5. 9=45 b 5 =b 1 q 4 = =5. 81=405 Answer:45; 405. Solution
UE3 1) Given: (a n), a 1 = – 3, and 2 = 4. Find: a 16 – ? 2) Given: (b n), b 12 = – 32, b 13 = – 16. Find: q – ? 3) Given: (a n), a 21 = – 44, and 22 = – 42. Find: d - ? 4) Given: (a n), a 1 = 28, and 21 = 4. Find: d - ? 5) Given: (b n), q = 2. Find: b 5 – ?
Assignments from the collection intended for preparation for the final certification in new form in algebra in grade 9, assignments are offered that are worth 2 points:) The fifth term of an arithmetic progression is equal to 8.4, and its tenth term is equal to 14.4. Find the fifteenth term of this progression) The number -3.8 is the eighth term of the arithmetic progression (a n), and the number -11 is its twelfth term. Is -30.8 a member of this progression? 6.5.1) Between the numbers 6 and 17, insert four numbers so that together with these numbers they form an arithmetic progression) In geometric progression b 12 = 3 15 and b 14 = 3 17. Find b 1.
Arithmetic and geometric progression What theme unites the concepts:
1) Difference 2) Sum n first terms 3) Denominator 4) First term
5) Arithmetic mean
6) Geometric mean?
Arithmetic
And
geometric
progression
Ustimkina L.I. Bolshebereznikovskaya secondary school
Progression Arithmetic Geometric
Ustimkina L.I. Bolshebereznikovskaya secondary school
The word progression comes from the Latin “progresio”.
So, progressio is translated as “moving forward.”
Ustimkina L.I. Bolshebereznikovskaya secondary school
The word progress is used in other fields of science, for example, in history, to characterize the process of development of society as a whole and the individual. In the presence of certain conditions any process can occur in both forward and reverse directions. The reverse direction is called regression, literally “moving backwards.”
Ustimkina L.I. Bolshebereznikovskaya secondary school
THE LEGEND ABOUT THE CREATOR OF CHESS
The first time on the control button, the second time on the sage
Ustimkina L.I. Bolshebereznikovskaya secondary school
Problem from the Unified State Exam The young man gave the girl 3 flowers on the first day, and on each subsequent day he gave 2 more flowers than on the previous day. How much money did he spend on flowers in two weeks if one flower costs 10 rubles?
224 flowers
224*10=2240 rub.
Ustimkina L.I. Bolshebereznikovskaya secondary school
http://uztest.ru
Complete tasks A6 and A1
Ustimkina L.I. Bolshebereznikovskaya secondary school
Eye exercise
Ustimkina L.I. Bolshebereznikovskaya secondary school
21-24 points - score “5”
17-20 points - score “4”
12-16 points – score “3”
0-11 points – score “2”
Ustimkina L.I. Bolshebereznikovskaya secondary school
Democritus
“ Good people become more from exercise than from nature”
Ustimkina L.I. Bolshebereznikovskaya secondary school
100,000 rub. for 1 kopeck
Ustimkina L.I. Bolshebereznikovskaya secondary school
100,000 for 1 kopeck
- The rich millionaire returned from his absence unusually joyful: he had a happy meeting on the road that promised great benefits.
- “There are such successes,” he told his family. “I met a stranger on the way, who didn’t show himself. And at the end of the conversation he offered such a profitable deal that took my breath away.
- “We’ll make this agreement with you,” he says. I will bring you a hundred thousand rubles every day for a whole month. Not without reason, of course, but the pay is trivial. On the first day, by agreement, I must pay - it’s funny to say - just one kopeck.
- One kopeck? - I ask again.
- “One kopeck,” he says. “For the second hundred thousand you’ll pay 2 kopecks.”
- Well, - I can’t wait. - And then?
- And then: for the third hundred thousand 4 kopecks, for the fourth 8, for the fifth - 16. And so on for a whole month, every day twice as much as the previous one.
Ustimkina L.I. Bolshebereznikovskaya secondary school
Received for
Gave
Received for
Gave
21st hundred
22nd hundred
10,485 rub. 76 kopecks.
20,971 rub. 52 kopecks.
23rd hundred
20,971 rub. 52 kopecks.
24th hundred
RUB 41,943 04 kop.
25th hundred
RUB 167,772 16 kopecks
26th hundred
RUR 335,544 32 kopecks
27th hundred
128 kopecks = 1 rub. 28 kopecks.
RUB 671,088 64 kopecks
10th hundred
28th hundred
RUB 1,342,177 28 kopecks
29th hundred
30th hundred
RUR 2,684,354 56 kopecks
RUB 5,368,709 12 kopecks
Ustimkina L.I. Bolshebereznikovskaya secondary school
The rich man gave: S 30
Given: b 1 =1; q=2; n=30.
S 30 =?
Solution
S n =
b 30 =1∙2 29 = 2 29
S 30 =2∙2 29 – 1= 2 ∙5,368,709 rub. 12 kop.–1 kop. =
= RUB 10,737,418 23 kopecks
RUB 10,737,418 23 kopecks - 3,000,000 rub. = RUR 7,737,418 23 kopecks – received by a stranger
Answer : RUB 10,737,418 23 kopecks
Ustimkina L.I. Bolshebereznikovskaya secondary school
1 slide
The 20th century has ended, but the term “progression” was introduced by the Roman author Boethius back in the 4th century. AD From the Latin word progressio - “moving forward”. The first ideas about arithmetic progression were among the ancient peoples. In cuneiform Babylonian tablets and Egyptian papyri there are progression problems and instructions on how to solve them. It was believed that the ancient Egyptian papyrus of Ahmes contained the oldest progression problem about rewarding the inventor of chess, dating back two thousand years. But there is a much older problem about dividing bread, which is recorded in the famous Egyptian Rhinda papyrus. This papyrus, discovered by Rind half a century ago, was compiled about 2000 BC and is a copy from another, even more ancient mathematical work, perhaps dating back to the third millennium BC. Among arithmetic, algebraic and geometric problems There is one of this document, which we present in free translation.
2 slide
12; 5; 8; 11;14; 17;… 2) 3; 9; 27; 81; 243;… 3) 1; 6; eleven; 20; 25;… 4) –4; -8; -16; –32; ... 5) 5; 25; 35; 45; 55;… 6) –2; -4; – 6; - 8; ... arithmetic progression d = 3 arithmetic progression d = – 2 geometric progression q = 3 sequence of numbers geometric progression q = 2 sequence of numbers
3 slide
4 slide
Studied this topic, The theory scheme has been completed, you have learned a lot of new formulas, and you have solved problems with progression. And now the beautiful slogan “PROGRESSIO - FORWARD” will lead us to the last lesson.
5 slide
Solution: Obviously, the amount of bread received by the participants in the section constitutes an increasing arithmetic progression. Let its first term be x, the difference be y. Then: a1 – Share of the first – x, a2 – Share of the second – x+y, a3 – Share of the third – x + 2y, a4 – Share of the fourth – x + 3y, a5 – Share of the fifth – x + 4y. Based on the conditions of the problem, we compose the following 2 equations:
6 slide
Problem 1: (problem from the Rind papyrus) One hundred measures of bread were divided between 5 people so that the second received as much more than the first as the third received more than the second, the fourth is greater than the third and the fifth is greater than the fourth. In addition, the first two received 7 times less than the other three. How much should you give each?
7 slide
8 slide
Slide 9
The lesson is over today, you couldn’t be more friendly. But everyone should know: Knowledge, perseverance, work will lead to progress in life.
10 slide
11 slide
Answers: 6.1 (20.4) (I) 6.2. (is), 6.5. (6;8.2;10’4;12’6;14’8;17.), 6.8. (b1=34 or b1= –34).
12 slide
Assignments from the collection intended for preparation for the final certification in the new form in algebra in grade 9, assignments are offered that are worth 2 points: 6.1. 1) The fifth term of an arithmetic progression is equal to 8.4, and its tenth term is equal to 14.4. Find the fifteenth term of this progression. 6.2. 1) The number –3.8 is the eighth member of the arithmetic progression (ap), and the number –11 is its twelfth member. Is -30.8 a member of this progression? 6.5. 1) Between numbers 6 and 17, insert four numbers so that together with these numbers they form an arithmetic progression. 6.8. 1) In geometric progression b12 = Z15 and b14 = Z17. Find b1.
Slide 13
Answers: 1) 102; (P) 2) 0.5; (B) 3) 2; (P) 4) 6; (D) 5) – 1.2; (E) 6) 8; (WITH)
Slide 14
"Carousel" - educational independent work 1) Given: (a n), a1 = – 3, a2 = 4. Find: a16 – ? 2) Given: (b n), b 12 = – 32, b 13 = – 16. Find: q – ? 3) Given: (a n), a21 = – 44, a22 = – 42. Find: d - ? 4) Given: (b n), bп > 0, b2 = 4, b4 = 9. Find: b3 – ? 5) Given: (a n), a1 = 28, a21 = 4. Find: d - ? 6) Given: (b n) , q = 2. Find: b5 – ? 7) Given: (a n), a7 = 16, a9 = 30. Find: a8 –? 1) (P) ;2) (V) ;3) (R); 4) (D); 5) (E); 6) (C).
15 slide
Properties of a geometric progression Given: (b n) geometric progression, b n >0 b4=6; b6=24 Find: b5 Solution: using the property of geometric progression we have: Answer: 12(D) Solution
16 slide
Properties of an arithmetic progression Given: (a n) arithmetic progression a4=12.5; a6=17.5 Find: a5 Solution: using the property of arithmetic progression we have: Answer: 15 (O) Solution
Slide 17
It is easy to see that the result is a magic square, the constant C of which is equal to 3a+12d. Indeed, the sum of the numbers in each row, in each column and along each diagonal of the square is equal to 3a + 12d. Let the following arithmetic progression be given: a, a+d, a+2d, a+3d, …, a+8d, where a and d are natural numbers. Let's arrange its members in a table.
18 slide
An interesting property of arithmetic progression. Now, let's look at another property of the members of an arithmetic progression. It will most likely be entertaining. We are given a “flock of nine numbers” 3, 5, 7, 9, 11, 13, 15,17, 19. It represents an arithmetic progression. In addition, this flock of numbers is attractive because it can fit into nine square cells so that a magic square is formed with a constant equal to 33
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Slide 1
Mathematics teacher Semyaninova E.N. MBOU "Voronezh Cadet School named after. A.V. Suvorov"
Slide 2
Playing the piano;
Only D. Polya can learn this.
Slide 3
The French word for dessert refers to sweet dishes served at the end of a meal. The names of some desserts, cakes and ice cream, are also of French origin. For example, ice cream “sundae” got its name from the French city of Plombières. Where it was first made according to a special recipe.
Slide 4
Find out the translation for the French word “meringue” (a light cake made from beaten egg whites and sugar)?
Slide 5
Slide 6
lightning - translation of the French word “éclair” (choux pastry with cream inside).
Slide 7
Progressions in life and everyday life
Slide 8
The vertical rods of the truss have the following length: the smallest is 5 dm, and each next one is 2 dm. longer. Find the length of seven such rods. Answer: 77 dm.
Slide 9
IN favorable conditions The bacterium multiplies so that it divides into three in 1 second. How many bacteria will be in the test tube after 5 seconds? Answer: 121
Slide 10
The truck transports a load of crushed stone weighing 210 tons, increasing the transportation rate by the same number of tons every day. It is known that 2 tons of crushed stone were transported on the first day. Determine how many tons of crushed stone were transported on the ninth day if all the work was completed in 14 days. 18 tons
Slide 11
A body falls from a tower 26 m high. In the first second it travels 2 m, and for each subsequent second it travels 3 m more than the previous one. How many seconds will it take the body to hit the ground? Answer: 4 seconds
Slide 12
For the first and last days The snail crawled a total of 10 meters. Determine how many days the snail spent on the entire journey if the distance between the trees is 150 meters. Answer: 30 days
Slide 13
A truck leaves point A at a speed of 40 km/h. At the same time, a second car set off from point B to meet him, which covered 20 km in the first hour, and each subsequent car covered 5 km more than the previous one. How many hours later will they meet if the distance from A to B is 125 km? Answer: 2 hours
Slide 14
The amphitheater consists of 10 rows, with each next row having 20 more seats than the previous one, and the last row having 280 seats. How many people can the amphitheater accommodate? Answer: 1900
Slide 15
A little history
Problems on geometric and arithmetic progressions are found among the Babylonians, in Egyptian papyri, and in the ancient Chinese treatise “Mathematics in 9 Books.”
Slide 16
Archimedes was the first to draw attention to the connection between progressions.
Slide 17
In 1544, the book “General Arithmetic” by the German mathematician M. Stiefel was published. Stiefel compiled the following table:
Slide 18
128 -3 7 -3+7=4 4 16 -4 -2 -1 0 1 2 3 5 6 64 6-(-1)=7 32 1 2 4 8
Slide 19
crossnumber
a b e c d g
Slide 20
5 1 1 2 1 1 2 6 5 0 0 5 0 0 8 1 3 a b c d e g
Slide 21
Problem solving
Slide 22
1. Solution: b2=3q, b3=3q2, q=-5; -4; -3; -2; -13; -15; 75 3; -12; 48;… 3; -9; 27;… 3; -6; 12;… 3; -3; 3;... Answer:
Slide 23
2. Three numbers form an arithmetic progression. If you add 8 to the first number, you get a geometric progression with a sum of terms of 26. Find these numbers. Solution: Answer: -6; 6; 18 or 10; 6; 2
Slide 24
3. An equation has roots, and an equation has roots. Determine k and m if the numbers are successive terms of an increasing geometric progression. hint Solution: - geometric progression Answer: k=2, m=32
Slide 25
Vieta's theorem: the sum of the roots of the reduced quadratic equation is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term.
Slide 26
literature
View all slides
Abstract
MBOU "Voronezh Cadet School"
school named after A.V. Suvorov"
Semyaninova E. N.
Problem solving is a practical art,
similar to swimming or skiing, or
imitating selected models and constantly training.
Find the sum of eleven terms of an arithmetic progression, the first term of which is equal to – 5, and the sixth is equal to – 3.5.
Answer: 77dm
Answer: 18 tons
Answer: 4 seconds
Snail
meters. (Slide 12)
Answer: 30 days
Answer: 1900
Another example.
64 6 -1 6 – (-1) = 7
It's not hard to figure out:
2-3∙ 27 = 24, 26: 2-1 = 27
V. Crossnumber. (Slide 19-20)
Work in groups.
Horizontally:
;
127; -119; …;
Vertically:
Given a geometric progression 3; b2; b3;…, whose denominator is an integer. Find this progression if
12q2 + 72q +35 =0
This means q=-5; -4; -3; -2; -1
Slide 4 | |||
Geometric progression |
Answer: -6; 6; 18 or 10; 6; 2
k And m
By Vieta's theorem
Numbers needed: 1; 2; 4; 8.
Answer: k= 2, m= 32
VII. Homework.
Solve problems.
Literature:
Algebra 9th grade. Tasks for the learning and development of students / comp. Belenkova E.Yu. "Intellect - Center". 2005.
Library of the magazine "Mathematics at School". Issue 23. Mathematics in puzzles, crosswords, chainwords, cryptograms. Khudadatova S.S. Moscow. 2003.
Mathematics. Supplement to the newspaper “First of September”. 2000. No. 46.
Multi-level didactic materials in algebra for 9th grade / comp. THOSE. Bondarenko. Voronezh. 2001.
MBOU "Voronezh Cadet School"
school named after A.V. Suvorov"
Semyaninova E. N.
Topic: Arithmetic and geometric progressions.
1) summarize information on progressions; improve the skills of finding the nth term and the sum of the first n terms of given progressions using formulas; solving problems that use both sequences;
2) continue the formation of practical skills;
3) develop the cognitive interest of students, teach them to see the connection between mathematics and the life around them.
Problem solving is a practical art,
similar to swimming or skiing, or
playing the piano; You can only learn this
imitating selected models and constantly training.
I. Organizing time. Explanation of lesson objectives. (Slide 2)
II. Warm up. In the world of interesting things. (Slide 3-6)
The French word for dessert refers to sweet dishes served at the end of a meal. The names of some desserts, cakes and ice cream, are also of French origin. For example, ice cream “plombier” got its name from the French city of Plombieres. Where it was first made according to a special recipe.
Using the answer you found and the table data, find out how the French word “meringue” (a light cake made from beaten egg whites and sugar) is translated?
Find the sum of eleven terms of an arithmetic progression, the first term of which is equal to – 5, and the sixth is equal to – 3.5.
The French word "meringue" means kiss. The second of the proposed words, “lightning,” is a translation of the French word “éclair” (a choux pastry with cream inside).
III. Progression in life and everyday life. (Slide 7)
Progression problems are not abstract formulas. They are taken from our life itself, connected with it and help solve some practical issues.
The vertical rods of the truss have the following length: the smallest is 5 dm, and each next one is 2 dm longer. Find the length of seven such rods. (Slide 8)
Answer: 77dm
Under favorable conditions, the bacterium multiplies so that it divides into three in 1 second. How many bacteria will be in the test tube after 5 seconds? (Slide 9)
The truck transports a load of crushed stone weighing 210 tons, increasing the transportation rate by the same number of tons every day. It is known that 2 tons of crushed stone were transported on the first day. Determine how many tons of crushed stone were transported on the ninth day if all the work was completed in 14 days. (Slide 10)
Answer: 18 tons
A body falls from a tower 6 m high. In the first second it travels 2 m, for each subsequent second it travels 3 m more than the previous one. How many seconds will it take the body to reach the ground? (Slide 11)
Answer: 4 seconds
A snail crawls from one tree to another. Every day she crawls the same distance further than the previous day. It is known that during the first and last days the snail crawled a total of 10 meters. Determine how many days the snail spent on the entire journey if the distance between the trees is 150
meters. (Slide 12)
Answer: 30 days
A truck leaves point A at a speed of 40 km/h. At the same time, a second car set off from point B to meet him, which covered 20 km in the first hour, and each subsequent car covered 5 km more than the previous one. How many hours later will they meet if the distance from A to B is 125 km? (Slide 13) Answer: 2 hours
The amphitheater consists of 10 rows, with each next row having 20 more seats than the previous one, and the last row having 280 seats. How many people can the amphitheater accommodate? (Slide 14)
Answer: 1900
IV. A little history. (Slide 15-16)
Problems on geometric and arithmetic progressions are found among the Babylonians, in Egyptian papyri, and in the ancient Chinese treatise “Mathematics in 9 Books.” Archimedes was apparently the first to draw attention to the connection between progressions. In 1544, the book “General Arithmetic” by the German mathematician M. Stiefel was published. Stiefel compiled the following table (Slide 17):
In the top line there is an arithmetic progression with a difference of 1. In the bottom line there is a geometric progression with a denominator of 2. They are arranged so that the zero of the arithmetic progression corresponds to the unit of the geometric progression. This is a very important fact.
Now imagine that we don’t know how to multiply and divide. It is necessary to multiply, for example, by 128. In the table above it is written -3, and above 128 it is written 7. Let's add these numbers. It turned out 4. Under 4 we read 16. This is the required product.
Another example.
Divide 64 by. We do the same:
64 6 -1 6 – (-1) = 7
The bottom line of the Stiefel table can be rewritten as follows:
2-4; 2-3; 2-2; 2-1; 20; 21; 22; 23; 24; 25; 26; 27.
It's not hard to figure out:
2-3∙ 27 = 24, 26: 2-1 = 27
We can say that if the exponents form an arithmetic progression, then the degrees themselves form a geometric progression. (Slide 18)
V. Crossnumber. (Slide 19-20)
Work in groups.
Crossnumber is one of the types of numerical puzzles. Translated from English word"crossnumber" means "cross number". When composing crossnumbers, the same principle is applied as when composing crosswords: one sign fits into each cell, “working” horizontally and vertically.
One number fits into each cell of the cross number (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). And to avoid confusion, task numbers are indicated by letters. The numbers to be guessed are only positive integers; recording such numbers cannot start from zero (i.e. 42 cannot be written as 042).
Some crossnumber questions may seem vague and allow for multiple (and sometimes very many) answers. But this is the style of crossnumbers. If they always gave only clear answers, then it wouldn't be a game.
Horizontally:
a) the number of odd numbers in the natural series, starting from 13, the sum of which is 3213;
c) the sum of the first five terms of a geometric progression, the fourth term of which is equal to 3, and the seventh is equal to ;
e) the sum of the first six positive terms of an arithmetic progression
127; -119; …;
e) the third term of a geometric progression (bn), in which the first term is 5 and the denominator g is 10;
g) the sum -13 + (-9) + (-5) + … + 63, if its terms are consecutive terms of an arithmetic progression.
Vertically:
A) the sum of all double digit numbers, multiples of nine;
B) double the twenty-first term of an arithmetic progression, in which the first term is equal to -5 and the difference is equal to 3;
B) the sixth term of the sequence, which is given by the formula of the nth term
D) the difference of an arithmetic progression, if.
VI. Solving non-standard problems. (Slide 21)
Given a geometric progression 3; b2; b3;…, whose denominator is an integer. Find this progression if
b2=3q, b3=3q2, then. Let's solve the inequality.
12q2 + 72q +35 =0
This means q=-5; -4; -3; -2; -1
Searched sequences: 3; -15; 75;…
Three numbers form an arithmetic progression. If you add 8 to the first number, you get a geometric progression with a sum of terms of 26. Find these numbers. (Slide 23).
B, c are the required numbers. Let's make a table.
Slide 4 | |||
Geometric progression |
By condition, the sum of three numbers forming a geometric progression is equal to 26, i.e. , в=6
We use the property of the terms of a geometric progression. We get the equation:
Answer: -6; 6; 18 or 10; 6; 2
An equation has roots, and an equation has roots. Define k And m, if the numbers are consecutive terms of an increasing geometric progression. (Slide 24-25)
Since the numbers form a geometric progression, we have:
By Vieta's theorem
We get, since the sequence is increasing.
Numbers needed: 1; 2; 4; 8.
Answer: k= 2, m= 32
VII. Homework.
Solve problems.
Find a geometric progression if the sum of its first three terms is 7 and their product is 8.
Divide the number 2912 into 6 parts so that the ratio of each part to the next is equal
In arithmetic progression it is and. How many terms of this progression must be taken so that their sum is 104?
Literature:
Algebra 9th grade. Tasks for the learning and development of students / comp. Belenkova E.Yu. "Intellect - Center". 2005.
Library of the magazine "Mathematics at School". Issue 23. Mathematics in puzzles, crosswords, chainwords, cryptograms. Khudadatova S.S. Moscow. 2003.
Mathematics. Supplement to the newspaper “First of September”. 2000. No. 46.
Multi-level didactic materials in algebra for grade 9/comp. THOSE. Bondarenko. Voronezh. 2001.
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