Lesson Problems on arithmetic progressions

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Problems on arithmetic progressions


This lesson presents some basic and typical problems on arithmetic progressions.

Problem 1

Derive the formula for the sum of the first  n  natural numbers.

Solution
The sequence of the first  n  natural numbers is  1, 2, 3, 4, 5, ..., n-1, n.
This is the arithmetic progression with the first term  a%5B1%5D+=+1  and the common difference  d+=+1.
Now apply the general formula for the sum of the first  n  terms of an arithmetic progression (see the formula  (2)  of the lesson  Arithmetic progressions  under
the current topic in this site):

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 = %281%2B%28n-1%29%2F2%29%2An = %282%2Bn-1%29%2F2%2An = %28n%2A%28n%2B1%29%29%2F2.

This is the formula for the sum of the first  n  natural numbers we are looking for.
Using this formula you can easily calculate the partial sums of the sequence of the natural numbers. These partial sums are
 1, 3, 6, 10, 15, 21, ...  for  n = 1, 2, 3, 4, 5, 6, ....


Problem 2

Derive the formula for the sum of the first  n   odd   positive integer numbers.

Solution
The sequence of the first  n  odd positive integer numbers is  1, 3, 5, 7, 9, ..., 2n-1.
This is the arithmetic progression with the first term  a%5B1%5D+=+1  and the common difference  d+=+2.
Let us calculate the partial sums of this progression.  You can easily do it yourself.  These partial sums are 1, 4, 9, 16, ... .  Do you see the pattern?
It looks like the sequence of the squares of sequential integer numbers, isn't?  You will see soon that it is true.

Apply the general formula for the sum of the first  n  terms of an arithmetic progression (see the formula  (2)  of the lesson  Arithmetic progressions  under
the current topic in this site):

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 = %281%2B%28%28n-1%29%2A2%29%2F2%29%2An = %281%2Bn-1%29%2An = n%5E2.

This is the formula for the sum of the first  n  odd positive integer numbers we are looking for.
Using this formula you can easily calculate the partial sums of the sequence of the odd positive integer numbers.  These partial sums are
 1, 4, 9, 16, 25, 36, ...  for  n = 1, 2, 3, 4, 5, 6,... .

It is the sequence of the squares of sequential integer numbers, indeed.  You just proved it!


Problem 3

Derive the formula for the sum of the first  n   even   positive integer numbers.

Solution
The sequence of the first  n  even positive integer numbers is  2, 4, 6, 8, 10, ..., 2n.
This is the arithmetic progression with the first term  a%5B1%5D+=+2  and the common difference  d+=+2.

Now, apply the general formula for the sum of the first  n  terms of an arithmetic progression (see the formula  (2)  of the lesson  Arithmetic progressions  under
the current topic in this site):

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 = %282%2B%28%28n-1%29%2A2%29%2F2%29%2An = %282%2Bn-1%29%2An = n%2A%28n%2B1%29.

This is the formula for the sum of the first  n  even positive integer numbers we are looking for.


Problem 4

Find the  21-th  term of the arithmetic progression if its  11-th  term is  53  and the  17-th  term is  71.

Solution
The major thing we need to do is to retrieve the  common difference  and the  first term  of the progression from the input data.  After that everything is simple.

OK, let us retrieve the common difference first.
Let us write the expressions for the  11-th  and  17-th terms.  Use the general formula for the  n-th  term of an arithmetical progression  (lesson Arithmetic progressions
under the current topic in this site)

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

and specify it for  n=11  and  n=17.  Use the given values for  a%5B11%5D  and  a%5B17%5D.  You get
a%5B11%5D = a%5B1%5D+%2B+10%2Ad = 53,
a%5B17%5D = a%5B1%5D+%2B+16%2Ad = 71.

In this way, you get the system of two linear equations
system+%28a%5B1%5D+%2B+10%2Ad+=+53%2C%0D%0Aa%5B1%5D+%2B+16%2Ad+=+71%29

with two unknowns a%5B1%5D and d.  To solve it, distract the first equation from the second one.  You get
16%2Ad-10%2Ad+=+71+-+53,   or

6d = 18.

Hence,  d=3.  Thus you got the  common difference  value  d=3.
Now, when you have retrieved the  common difference  d,  you can find the first term of the progression.  For this use, for instance, the equation

a%5B1%5D+%2B+10%2Ad+=+53,

which you just got above.  From this equation

a%5B1%5D = 53+-+10%2Ad = 53-10%2A3 = 53+-+30 = 23.

So, our progression is 23, 26, 29, ... and so on. Check that its 11-th and 17-th terms are

a%5B11%5D = a%5B1%5D+%2B+%28n-1%29%2Ad = 23+%2B+%2811-1%29%2A3 = 23%2B10%2A3 = 23+%2B+30 = 53 and
a%5B17%5D = a%5B1%5D+%2B+%28n-1%29%2Ad = 23+%2B+%2817-1%29%2A3 = 23%2B16%2A3 = 23+%2B+48 = 71

and agree with the input data. Now the 21-th term of the progression is

a%5B21%5D = a%5B1%5D+%2B+%28n-1%29%2Ad = 23+%2B+%2821-1%29%2A3 = 23%2B20%2A3 = 23+%2B+60 = 83.

The problem is solved. The answer is a%5B21%5D = 83.


Problem 5

Find the sum of the first  31  terms of the arithmetic progression if its  7-th  term is equal to  21  and the  11-th  term is equal to  29.

Solution
This problem is similar to the preceding one (Problem 4).
Again, the major thing we need to do is to retrieve the  common difference  and the  first term  of the progression from the input data.  After that everything is simple.

OK, let us retrieve the common difference first.
Let us write the expressions for the  7-th  and  11-th terms.  Use the general formula for the  n-th  term of an arithmetical progression  (lesson Arithmetic progressions
under the current topic in this site)

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

and specify it for  n=7  and  n=11.  Use the given values for  a%5B7%5D  and  a%5B11%5D.  You get
a%5B7%5D = a%5B1%5D+%2B+6%2Ad = 21,
a%5B11%5D = a%5B1%5D+%2B+10%2Ad = 29.

In this way, you get the system of two linear equations
system+%28a%5B1%5D+%2B+6%2Ad+=+21%2C%0D%0Aa%5B1%5D+%2B+10%2Ad+=+29%29

with two unknowns a%5B1%5D and d.  To solve it, distract the first equation from the second one.  You get
10%2Ad-6%2Ad+=+29+-+21,   or

4d = 8.

Hence,  d=2.  Thus you got the  common difference  value  d=2.
Now, when you have retrieved the  common difference  d,  you can find the first term of the progression.  For this use, for instance, the equation

a%5B1%5D+%2B+6%2Ad+=+21,

which you just got above.  From this equation

a%5B1%5D = 21+-+6%2Ad = 21-6%2A2 = 21+-+12 = 9.

So, our progression is 9, 11, 13, ... and so on. Check that its 7-th and 11-th terms are

a%5B7%5D = a%5B1%5D+%2B+%28n-1%29%2Ad = 9+%2B+%287-1%29%2A2 = 9%2B6%2A2 = 9+%2B+12 = 21 and
a%5B11%5D = a%5B1%5D+%2B+%28n-1%29%2Ad = 9+%2B+%2811-1%29%2A2 = 9%2B10%2A2 = 9+%2B+20 = 29

and agree with the input data. Now the sum of the first 31-th terms of the progression is equal to

S%5B31%5D = %28a%5B1%5D+%2B+%28n-1%29%2Ad%2F2%29%2An = %289+%2B+%2831-1%29%2A2%2F2%29%2A31 = %289%2B30%29%2A31 = 39%2A31 = 1209.

We used here the formula for the sum of the first  n   (n=31)  terms of an arithmetic progression  (lesson  Arithmetic progressions  under the current topic in this site).

The problem is solved. The answer is S%5B31%5D = 1209.


For more word problems on arithmetic progressions see the lesson Word problems on arithmetic progressions under the current topic in this site.


My lessons on arithmetic progressions in this site are
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions                                                                                          (this lesson)
    - 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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