Lesson Arithmetic progressions

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Arithmetic progressions


Definition
An  arithmetic progression  is a sequence of numbers such that the difference between the current term and the preceding term is the same
for any two consecutive terms.
The constant value of the difference between the current and the preceding consecutive terms of the arithmetic progression is called the  common difference
of the arithmetic progression.


Examples


1. The sequence of the numbers
2, 5, 8, 11, 14, 17
is the arithmetic progression because the difference between the current term and the preceding term is equal to 3 for any two consecutive terms.
The first term of this progression is equal to 2, the common difference is equal to 3.
We can present this progression as a sequence: a%5B1%5D=2, a%5B2%5D=5, a%5B3%5D=8, a%5B4%5D=11, a%5B5%5D=14, a%5B6%5D=17. Below we will use an index n to designate the current term of a sequence.
For instance, if n=3, a%5Bn%5D+=+a%5B3%5D+=+8 is the third term of the progression.

2. The sequence of all the odd integer positive numbers
1, 3, 5, 7, 9, 11, 13, ...
is the arithmetic progression because the difference between the current term and the preceding term is equal to 2 for any two consecutive terms.
The first term of this progression is equal to 1, the common difference is equal to 2. This progression is an infinite sequence.

3. The sequence of all the even integer positive numbers
2, 4, 6, 8, 10, 12, 14, ...
is the arithmetic progression because the difference between the current term and the preceding term is equal to 2 for any two consecutive terms.
The first term of this progression is equal to 2, the common difference is equal to 2. This progression is an infinite sequence.

4. The sequence of numbers
1.5, 2.4, 3.3, 4.2, 5.1, 6.0, 6.9, 7.8
is the arithmetic progression. The difference between the current term and the preceding term is the constant value of 0.9 for any two consecutive terms.
The first term of this progression is equal to 1.5, the common difference is equal to 0.9.

5. The sequence of numbers
1.5, 0.5, -0.5, -1.5, -2.5, -3.5, -4.5, -5.5
is the arithmetic progression. The difference between the current term and the preceding term is the constant value of -1 for any two consecutive terms.
The first term of this progression is equal to 1.5, the common difference is equal to -1. This progression is a descending sequence.


Formula for n-th term of an arithmetic progression

If  a%5B1%5D, a%5B2%5D, a%5B3%5D, ..., a%5Bn%5D  is the arithmetic progression with the first term  a%5B1%5D  and the common difference  d  then the formula for the  n-th term is

      a%5Bn%5D = a%5B1%5D+%2B+%28n-1%29%2Ad.                                           (1)


For the proof of the formula (1) see the lesson
    The proofs of the formulas for arithmetic progressions
under the current topic in this site.


Examples


6. Find the 25-th term of the arithmetic progression if the first term is 7 and the common difference is 12.

Solution.
Apply the formula (1) for the n-th term of an arithmetic progression. You have

a%5B25%5D = a%5B1%5D+%2B+%28n-1%29%2Ad = 7+%2B%2825-1%29%2A12 = 7%2B24%2A12 = 7%2B288 = 295.


7. Find the first term of the arithmetic progression if its 25-th term is equal to 72 and the common difference is 7.

Solution.
Apply the formula (1) for the n-th (25-th) term of an arithmetic progression. You have

a%5B25%5D = a%5B1%5D+%2B+%28n-1%29%2Ad = a%5B1%5D+%2B+%2825-1%29%2A7.

So, you get the equation for the unknown first term a%5B1%5D:
a%5B1%5D+%2B+%2825-1%29%2A7 = 72.

From this equation you have

a%5B1%5D = 72+-+%2825-1%29%2A7 = 72+-+24%2A7 = 72+-+168 = -96.

Thus a%5B1%5D = -96.
You can check this answer by substituting the found value of  a%5B1%5D  into the formula (1) for the 25-th term of the arithmetic progression:
a%5B25%5D = a%5B1%5D+%2B+%28n-1%29%2Ad = -96+%2B+%2825-1%29%2A7 = -96+%2B+24%2A7 = -96%2B168 = 72.

This check shows that the answer  a%5B1%5D = -96  is correct.
Thus the arithmetic progression you are looking for is -96, -89, -82, -75, ..., 65, 72. The number 72 is the 25-th term of the progression.


Formula for sum of an arithmetic progression

If  a%5B1%5D, a%5B2%5D, a%5B3%5D, ..., a%5Bn%5D  is the arithmetic progression with the first term  a%5B1%5D  and the common difference  d  then the formula for the sum of its first  n  terms is

      S%5Bn%5D = .        (2)

Or, the equivalent formula is
      S%5Bn%5D = %28a%5B1%5D+%2B+a%5Bn%5D%29%2F2%2An.                                           (3)


For the proof of the formulas (2) and (3) see the lesson
    The proofs of the formulas for arithmetic progressions
under the current topic in this site.


Examples


8. Find the sum of the first 25 natural numbers  1+%2B+2+%2B+3+%2B+ellipsis+%2B+25.

Solution.
The sequence  1, 2, 3, 4, ..., 25  is the arithmetic progression. The first term is equal to 1 and the difference is 1.
Now, apply the formula (2) for the sum of the first n terms of an arithmetic progression. You have

S%5B25%5D = 1+%2B+2+%2B+3+%2B+ellipsis+%2B+25 = %28a%5B1%5D+%2B+%28n-1%29%2Ad%2F2%29%2An = %281+%2B%2825-1%29%2A1%2F2%29%2A25 = %281%2B24%2F2%29%2A25 = %281%2B12%29%2A25 = 13%2A25 = 325.

If you apply the formula (3), you will get the same answer:

S%5B25%5D = 1+%2B+2+%2B+3+%2B+ellipsis+%2B+25 = %28%28a%5B1%5D+%2B+a%5Bn%5D%29%2F2%29%2An = %28%281+%2B25%29%2F2%29%2A25 = 13%2A25 = 325.


9. Find the sum of the first 50 natural numbers  1+%2B+2+%2B+3+%2B+ellipsis+%2B+50.

Solution.
The sequence  1, 2, 3, 4, ..., 50  is the arithmetic progression. The first term is equal to 1 and the difference is 1.
Now, apply the formula (2) for the sum of the first n terms of an arithmetic progression. You have

S%5B50%5D = 1+%2B+2+%2B+3+%2B+ellipsis+%2B+50 = %28a%5B1%5D+%2B+%28n-1%29%2Ad%2F2%29%2An = %281+%2B%2850-1%29%2A1%2F2%29%2A50 = %281%2B49%2F2%29%2A50 = %281%2B24.5%29%2A50 = 25.5%2A50 = 1275.

If you apply the formula (3), you will get the same answer:

S%5B50%5D = 1+%2B+2+%2B+3+%2B+ellipsis+%2B+50 = %28%28a%5B1%5D+%2B+a%5Bn%5D%29%2F2%29%2An = %28%281+%2B50%29%2F2%29%2A50 = %2851%2F2%29%2A50 = 25.5%2A50 = 1275.


10. Find the sum of the first 100 natural numbers  1+%2B+2+%2B+3+%2B+ellipsis+%2B+99+%2B+100.

Solution.
The sequence  1, 2, 3, 4, ..., 99, 100  is the arithmetic progression. The first term is equal to 1 and the difference is 1.
Now, apply the formula (2) for the sum of the first n terms of an arithmetic progression. You have

S%5B100%5D = 1+%2B+2+%2B+3+%2B+ellipsis+%2B+99+%2B+100 = %28a%5B1%5D+%2B+%28n-1%29%2Ad%2F2%29%2An = %281+%2B%28100-1%29%2A1%2F2%29%2A100 = %281%2B99%2F2%29%2A100 = %281%2B49.5%29%2A100 = 50.5%2A100 = 5050.

If you apply the formula (3), you will get the same answer:

S%5B100%5D = 1+%2B+2+%2B+3+%2B+ellipsis+%2B+100 = %28%28a%5B1%5D+%2B+a%5Bn%5D%29%2F2%29%2An = %28%281+%2B100%29%2F2%29%2A100 = %28101%2F2%29%2A100 = 50.5%2A100 = 5050.

Formula for sum of the first n natural numbers


The sequence of the first  n  natural numbers  1, 2, 3, . . . ,n  is the specific arithmetic progression.  There is a compact convenient general formula to calculate its sum.
Indeed, the first term of this progression is  1,  the common difference is  1,  and the number of terms is  n.  Substitute these values to the formula  (2),  and you get
the formula for the sum of the first  n  natural numbers

S%5Bn%5D = %28a%5B1%5D+%2B+%28%28n-1%29%2Ad%29%2F2%29%2An = %281+%2B+%28%28n-1%29%2A1%29%2F2%29%2An = %282+%2B+n-1%29%2An%2F2 = n%2A%28n%2B1%29%2F2.

Or, substitute the first term  a%5B1%5D+=+1  and the last term  a%5Bn%5D+=+n  into the formula (3). You get the same formula for the sum of the first  n  natural numbers
S%5Bn%5D = %28%281+%2B+n%29%2F2%29%2An = n%2A%28n%2B1%29%2F2.

Note that examples 8, 9 and 10 for the sums of the first 25, 50 and 100 first natural numbers agree with this formula:

S%5B25%5D = 25%2A26%2F2 = 25%2A13 = 325,
S%5B50%5D = 50%2A51%2F2 = 25%2A51 = 1275,
S%5B100%5D = 100%2A101%2F2 = 50%2A101 = 5050.


Summary


An arithmetic progression is a sequence of numbers with a constant difference between the current term and the preceding term for any two consecutive terms.

The n-th term of the arithmetic progression  a%5B1%5D, a%5B2%5D, a%5B3%5D, ..., a%5Bn%5D  with the first term  a%5B1%5D  and the common difference  d  is
    a%5Bn%5D = a%5B1%5D+%2B+%28n-1%29%2Ad.

The sum of the first n terms of the arithmetic progression  a%5B1%5D, a%5B2%5D, a%5B3%5D, ..., a%5Bn%5D  with the first term  a%5B1%5D  and the common difference  d  is equal to
    S%5Bn%5D = a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+%2B+ellipsis+%2B+a%5Bn%5D = %28a%5B1%5D+%2B+%28n-1%29%2Ad%2F2%29%2An.

Another (equivalent) formula for the sum of the first n terms of an arithmetic progression  a%5B1%5D, a%5B2%5D, a%5B3%5D, ..., a%5Bn%5D  is
    S%5Bn%5D = %28a%5B1%5D+%2B+a%5Bn%5D%29%2F2%2An.

The formula for the sum of the first  n  natural numbers is

1+%2B+2+%2B+3+%2B+...+%2B+n = %28%281+%2B+n%29%2F2%29%2An = n%2A%28n%2B1%29%2F2.


For some basic and typical problems on arithmetic progressions see the lesson Problems on arithmetic progressions under the current topic in this site.
For more word problems on arithmetic progressions see the lesson Word problems on arithmetic progressions under the current topic in this site.


The full list of my lessons on arithmetic progressions in this site is
    - Arithmetic progressions                                                                                                            (this lesson)
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
    - Chocolate bars and arithmetic progressions
    - Free fall and arithmetic progressions
    - Uniformly accelerated motions and arithmetic progressions
    - Increments of a quadratic function form an arithmetic progression
    - One characteristic property of arithmetic progressions
    - Solved problems on arithmetic progressions
    - Calculating partial sums of arithmetic progressions
    - Finding number of terms of an arithmetic progression
    - Inserting arithmetic means between given numbers
    - Advanced problems on arithmetic progressions
    - Interior angles of a polygon and Arithmetic progression
    - Math Olympiad level problems on arithmetic progression
    - Problems on arithmetic progressions solved MENTALLY
    - Entertainment problems on arithmetic progressions
    - Mathematical induction and arithmetic progressions
    - Mathematical induction for sequences other than arithmetic or geometric

OVERVIEW of my lessons on arithmetic progressions with short annotations is in the lesson  OVERVIEW of lessons on arithmetic progressions.

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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