An arithmetic progression is a sequence of numbers with special properties. And often appear in math problems. Below will provide the necessary information for you about the addition, properties and common formulas.
1. The concept of the additive level? What is wrong? For example?
Levels plus is a sequence of numbers (finite or infinite) of terms satisfying the condition from the 2nd term onwards equal to the term before it plus a constant number.
That constant term is called . Public wrong.
Recipe:
Un = Un-1 + d (n>=2)
For example:
– A constant sequence with constant terms is an exponent plus the common difference equal to O.
– The range of natural numbers 2; 4; 6; 8; 10;… is an exponent plus the difference of 2.
2. Additive properties?
If it is an arithmetic progression, then from the second term, every term (except the last term for a finite arithmetic progression) is the average of the two terms next to it in the sequence.
Recipe:
Un = (Un-1 + Un+1) : 2
Wallet example:
We have 3 consecutive terms of an arithmetic progression: 10; twelfth; 14
Then (10+14):2 = 1
3. Calculating the difference of arithmetic progression?
The number d is called the work difference of the arithmetic progression. Then the formula for calculating fairness is equal to:
Recipe:
d=Un+1 – Un
For example:
We have the sequence 1, 5, 9, 13, 17, 21, 25 which is an exponent plus the difference d = 4
because: 25-21=4; 21-17=4;…
4. General terms of the arithmetic progression:
If the starting additive is an element and the difference is d, then the nth term of the additive is calculated by arithmetic progression formula after:
Un=Ufirst+ (n-1)d
For example:
The arithmetic progression is 5,9,13;… n. know a sequence of 7 numbers.
Then: the nth term is equal to: 5 + 6.4 = 29.
5. Some other formulas:
5.1. Relation formula between any two terms
Un=Um + (nm)d
5.2. Formula to sum first n terms (nth partial sum) through first term and nth term
Sn = Ufirst+U2+ …+ Un = (n(U1+Un)/2)
6. Some types of exercises to calculate addition:
6.1. Form 1: Recognizing arithmetic progressions
Step 1: Find the common difference when knowing two consecutive terms according to the formula: d=un–un–1,∀n≥2.
Step 2: Conclude:
If d is a constant number, the sequence (un) is CSC.
If d varies with n, then the sequence (un) is not a CSC.
Wallet example: Given the following sequence of numbers: 3; 5; 7; 9; 13. Is the above sequence an additive number?
The variance of the above series is: 5-3=2; 7-5=2; 13-9=4.
Due to wrongdoing, there was a change.
Therefore, the sequence above is not an arithmetic progression.
6.2. Form 2: Find the difference from the arithmetic progression formula:
Example: Given an arithmetic progression (Un) that has U1=1 and the sum of the first 100 terms is 24850. What is the difference?
I have WILL100 = 24850
(n(1+ Un)/2) = 24850
U100 = 496
So U100 = 1= 99d
d= (24850-1)/99
d=5
6.3. Form 3: Find the terms of the arithmetic progression:
For the arithmetic progression Un have Ufirst = 5, d = 4 . Let’s calculate U26
We have :
U26 = Ufirst + (26 – 1) d
= 5 + (26 – 1) x 4
=105
6.4. Form 4: Calculate the arithmetic sum of the first n terms:
For example: An arithmetic progression (un) know that the first term ufirst = 5, the 11th term is u11 = 25. Sum the first 11 terms of this sequence?
Apply the formula Sn=(u1+un)n2
ufirst= 5
u11= 25
n = 11
Based on the above formula, we calculate the sum of the first 11 terms: Sn=(5+25)2.11=165
6.5. Form 5: Find the arithmetic progression:
Making:
Find the factors determining an arithmetic progression such as: first term u1, difference d.
Find the formula for the general term un=u1+(n–1)d
Wallet example: Determine the arithmetic progression such that the sum of the first n terms is equal to n=1 times half of the second term
7. Some example exercises:
Question 1: Prove the sequence of numbers (un) with un = 17n + 2 is the arithmetic progression
Direction Detailed explanation:
I have: un+1 = 17(n + 1) + 2 = 17n + 19
=> Brand: un+1 – un = (17n + 19) − (17n + 2) = 17
Derived: (un) is the arithmetic progression plus the difference d = 17.
Verse 2: Given the arithmetic progression (un)
a) (un) has a general term of: un= 7n – 3. Calculate100.
b) (un) have u2+ u22 = 40. Calculation
c) (un) have u4 + u8+ utwelfth + u16 = 224. Calculation19.
Direction Detailed explanation:
a) From the general term formula
We have:
First term: ufirst = 7 . 1 – 3 = 4;
The second term is : u2 = 7 . 2 – 3 = 11;
Common difference: d = 11 – 4 = 7
Then we have:
S100=n2u1+(n−1)d2=100[2.4+(100−1).7]2=35050
b) We have: u2+u22=40⇔u1+d+u1+21d=40⇔2u1+22d=40
So S23=232u1+22d2=23.402=460.
c) We have: u4+ u8 + utwelfth+ u16 = 224
⇔u1+3d+u1+7d+u1+15d=224⇔4u1+36d=224⇔u1+9d=56
So S19=192u1+18d2=19u1+9d=19.56=1064.
Sentence 3: Given the sequence of numbers (un) with un = 2n + 3. Prove that the sequence (un) is not an arithmetic progression.
Direction Detailed explanation:
I have: un+1 = 2n+1 + 3
Test signal: un+1 – un = (2 .)n+1 + 3) − (2n + 1)= 2n+1 – 2n
=> (un+1 – un) is not a constant; also depends on n. So the sequence of numbers (un) is not an arithmetic progression.
Sentence 4: Prove that:
a) If three numbers a, b, c form an additive, then the three numbers x, y, z also form an additive, with: x = a2– bc, y = b2– ca, z = c2 – ab.
b) If the equation x3– ax2+ bx – c = 0 has three solutions that form an additive, then 9ab = 2a3 + 27c.
Direction Detailed explanation:
a) a, b, c are additive so a + c = 2b
It is necessary to prove that x, y, z also form an additive ie x + z = 2y.
We have 2y = 2b2 – 2ca
And x + z = a2 + c2 – b(a + c)
= (a + c)2 – 2ac – 2b2
= 4b2 – 2ac – 2b2
= 2b2 – 2ac = 2y
Then we get: y=x+z2y=x+z2
So we have something to prove.
b) Suppose the equation has three solutions xfirstx2x3form an arithmetic progression then: xfirst+ x3 = 2x2 (first)
On the other hand: x3 – ax2 + bx – c = (x – xfirst)(x – x2)(x – x3)
= x3 – (xfirst + x2 + x3)x2 + (xfirst x2 + x2 x3 + x3 xfirst)x – xfirst x2 x3
Deduce xfirst + x2 + x3 = a (2)
From (1) and (2), we get 3×2=a⇔x2=a33x2=a⇔x2=a3
Since the given equation has a solution x2=a3x2=a3, that is:
(a3)3−a(a3)2+b(a3)−c=0⇔−2a327+ba3−c=0⇔9ab=2a3+27ca33−aa32+ba3−c=0⇔−2a327+ba3−c= 0⇔9ab=2a3+27c
So we have something to prove.
Sentence 5: Calculate the following sums:
a) S = 1 + 3 + 5 +… + (2n – 1) + (2n + 1)
b) S = 1 + 4 + 7 +… + (3n – 2) + (3n + 1) + (3n + 4)
c) S = 1002– 992+ 982 – 972 +… + 22 – first2
Direction Detailed explanation:
a) We have a sequence of numbers 1;3;5;…;(2n – 1);(2n + 1) is an arithmetic progression plus the difference d = 2 and ufirst = 1, general term uk= ufirst+ (k – 1)d.
We check what number 2n + 1 is in the sequence: 2n + 1 = ufirst + (k – 1)d
. Thus the sequence has n + 1 terms.
So .
b) We have the sequence number 1; 4; 7; … (3n – 2);(3n + 1);(3n + 4) is the exponent plus the difference d = 3 and ufirst= 1, general term uk= ufirst + (k – 1)d.
We check what number 2n + 1 is in the sequence: 3n + 4 = ufirst + (k – 1)d
. Thus the sequence has n + 2 terms.
So .
c) S = 1002– 992 + 982– 972 +… + 22 – first2
= (100 – 99)(100 + 99) + (98 – 97)(98 + 97) +… + (2 – 1)(2 + 1)
= 199 + 195 +… + 3
= 3 + 7 +… + 195 + 199
We have the sequence number 3; 7; …195; 199 is the arithmetic progression plus the difference d = 4, the first term ufirst = 3 and the nth term is un = 199.
Hence there is 199=3+n−1.4⇒n=50.
So .
Sentence 6: Solved problem: A factory has posted to recruit workers with salary compensation as follows: In the first quarter, the factory pays 6 million VND/quarter and from the second quarter will increase to 0.5 million for 1 precious. With the above treatment, after 5 years of working at the factory, what is the total salary of that worker?
Direction Detailed explanation:
Suppose workers work for the factory for n quarters, then the salary is denoted (un) (million dong)
By subject:
First quarter: ufirst = 6
Next quarters: un+1 = un + 0.5 with n 1
The worker’s salary per quarter is a term of the sequence un. On the other hand, the salary of the following quarter is 0.5 million more than the salary of the previous quarter, so the sequence un is an exponent plus the common difference d = 0.5.
We know 1 year will have 4 quarters => 5 years will have 5.4 = 20 quarters. According to y/c of the problem, we need to calculate the sum of the first 20 terms of the arithmetic progression (un).
20th quarter salary of workers: u20 = 6 + (20 – 1).0.5 = 15.5 million dong
Total salary of workers received after 5 years working at the factory: 215 (million VND)
Verse 7: An arithmetic progression (un) know that the first term ufirst = 5, the 11th term is u11 = 25. Sum the first 11 terms of this sequence
Detailed explanation instructions:
Applying formula Sn=(ufirst+un)n2
Based on the above formula, we calculate the sum of the first 11 terms: Sn=(5+25)2.11=165
Verse 8: Given 1 arithmetic progression (un) have ufirst = 1 and the difference d = 2. The sum of the first 3 terms of this arithmetic progression is
A. 5
B. 8
C. 9
D. 12
Detailed explanation instructions:
Applying formula: Sn=2ufirst+d(n–first)2n
Based on the above formula, we calculate the sum of the first 3 terms: Sn=2.1+2(3–first)2.3=9
Choose answer C.
Verse 9: A factory has posted to recruit workers with salary compensation as follows: In the first quarter, the factory pays VND 6 million/quarter and from the second quarter will increase to 0.5 million for 1 quarter. With the above treatment, after 5 years of working at the factory, what is the total salary of that worker?
A. 215 million won
B. 15.5 million won
C. 155 million won
D. 60 million
Detailed explanation instructions:
Suppose workers work for the factory for n quarters, then the salary is denoted (un) (million dong)
By subject:
- First quarter: ufirst = 6
- Next quarters: un+1 = un + 0.5 with n 1
The worker’s salary per quarter is a term of the series un. On the other hand, the salary of the following quarter is 0.5 million more than the salary of the previous quarter, so the sequence un is an exponent plus the common difference d = 0.5.
We know 1 year will have 4 quarters => 5 years will have 5.4 = 20 quarters. According to the problem’s y/c, we need to calculate the sum of the first 20 terms of the arithmetic progression (un).
20th quarter salary of workers: u20 = 6 + (20 – 1).0.5 = 15.5 million dong
Total salary of workers received after 5 years working at the factory: Stwelfth=20.(6+15,5)2=215 (million dong)
Choose answer A.
