Lesson Solved problems on arithmetic progressions

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Solved problems on arithmetic progressions


Problem 1

An arithmetic progression consists of three terms whose sum is  48  and the sum of their squares is  800.  Find the progression.

Solution 

One can present the three consecutive terms of the AP as

x -d, x, x + d,

where x is the mid term and d is the common difference.

Then the sum of the tree terms is 3x, and you can easily find a from the equation

3x = 48,

which implies x = 48%2F3 = 16.

Now the sum of squares of the tree terms is

%2816-d%29%5E2+%2B+16%5E2+%2B+%2816%2Bd%29%5E2 = %28256+-+32d+%2B+d%5E2%29+%2B+256+%2B+%28256+%2B+32d+%2B+d%5E2%29 = 3%2A256+%2B+2d%5E2 = 768+%2B+2d%5E2.

It gives you an equation to find d:

768+%2B+2d%5E2 = 800  --->  2d%5E2 = 800+-+768 = 32  --->  d%5E2 = 32%2F2 = 16.

Hence, d = +/- 4.

It gives the AP terms as  12, 16 20,   or   20, 16, 12.


Answer. AP terms are  12, 16 20,   or   20, 16, 12.

Problem 2

If the  6-th term of an arithmetic progression is  121,  find the sum of the first  11  terms.

Solution 

We can write 

a%5B7%5D  = a%5B6%5D%2Bd,   a%5B5%5D = a%5B6%5D-d,

a%5B8%5D  = a%5B6%5D%2B2d,  a%5B4%5D = a%5B6%5D-2d,

a%5B9%5D  = a%5B6%5D%2B3d,  a%5B3%5D = a%5B6%5D-3d,

a%5B10%5D = a%5B6%5D%2B4d,  a%5B2%5D = a%5B6%5D-4d,

a%5B11%5D = a%5B6%5D%2B5d,  a%5B1%5D = a%5B6%5D-5d.


Then 

 = 10%2Aa%5B6%5D.

Add a%5B6%5D, and you will get that the sum of the first 11 terms is equal to 11a%5B6%5D.

Hence, this sum is 11*21 = 231.


Answer.  231.

Problem 3

The sum of the first  5  terms of an AP is  30  and the fourth term is  44.  Find the common difference and the sum of the first  10 terms.

Solution 

Let "x" be the THIRD term and "d" be the common difference. 

Then the first 5 terms of the progression are

x - 2d, x-d, x, x+d, x + 2d,

and their sum is 5x.

Since the sum of the first 5 terms of the AP is 30, you have this equation

5x = 30,  which implies x = 6.

So, you just found the third term. It is 6.

Now, since the 4-th term is 44, the common difference is 44-6 = 38.

Then the second term is 6-38 = -32 and the first term is -32 - 38 = -70.

So, you know all and everything about the progression: The first term is -70 and the common difference is 38.

Then the sum of the first 10 terms of the progression is

S = -70+%2B+%28%289%2A6%29%2F2%29%2A10 = -70+%2B+270 = 200.

Problem 4

If the fifth term of arithmetic sequence is  23  and the sum of the first ten terms of the sequence is  240,
then find the sum of the first sixty terms of this sequence.

Solution 

Notice that

   =

= %28a%5B1%5D%2Ba%5B10%5D%29 + %28a%5B2%5D%2Ba%5B9%5D%29 + %28a%5B3%5D%2Ba%5B8%5D%29 + %28a%5B4%5D%2Ba%5B7%5D%29 + %28a%5B5%5D%2Ba%5B6%5D%29.


Also notice that all 5 the sums in pairs are the same.

So, you actually have  5%2A%28a%5B5%5D%2Ba%5B6%5D%29 = 240,  which implies

a%5B5%5D+%2B+a%5B6%5D = 240%2F5 = 48,  and then

23+%2B+a%5B6%5D = 48,  which gives  a%5B6%5D = 48 - 23 = 25.


Thus the common difference is  d = a%5B6%5D-a%5B5%5D = 25-23 = 2.


Then  a%5B1%5D = a%5B5%5D-4%2Ad = 23 - 4*2 = 23 - 8 = 15.


So, the progression has the first term of 15 and the common difference of 2.


Then a%5B60%5D = 15 + 2*59 = 133   and the sum of the first 60 terms is


S%5B60%5D = %28%28a%5B1%5D%2Ba%5B60%5D%29%2F2%29%2A60 = %28%2815%2B133%29%2F2%29%2A60 = (15+133)*30 = 4440.


Answer.  S%5B60%5D = 4440.

Problem 5

The sum of the fifth and seventh terms of an arithmetic series is  38,  while the sum of the first fifteen terms is  375.
Determine the first term and the common difference.

Solution 

1.  %28a%5B5%5D%2Ba%5B7%5D%29%2F2 = a%5B6%5D.   (1)


    Indeed, a%5B5%5D = a%5B6%5D+-d+,
            a%5B7%5D = a%5B6%5D+%2B+d.

    Adding, you get (1).

    It is a general (and characteristic) property of any arithmetic progression:


            for any three consecutive terms  a%5Bk-l%5D,  a%5Bk%5D and  a%5Bk%2Bl%5D  the middle term is equal to the half-sum of its neighbors.

            See the lesson One characteristic property of arithmetic progressions in this site.


2.  a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+%2B+ellipsis+%2B+a%5B15%5D = 15%2Aa%5B8%5D.


    It is easy to prove expressing the terms as neighbors of the central term a%5B8%5D:

    a%5B7%5D = a%5B8%5D-d,    a%5B9%5D = a%5B8%5D%2Bd,
    a%5B6%5D = a%5B8%5D-2d,   a%5B9%5D = a%5B8%5D%2B2d,
    a%5B5%5D = a%5B8%5D-3d,   a%5B9%5D = a%5B8%5D%2B3d,
    a%5B4%5D = a%5B8%5D-4d,   a%5B9%5D = a%5B8%5D%2B4d,
    a%5B3%5D = a%5B8%5D-5d,   a%5B9%5D = a%5B8%5D%2B5d,
    a%5B2%5D = a%5B8%5D-6d,   a%5B9%5D = a%5B8%5D%2B6d,
    a%5B1%5D = a%5B8%5D-7d,   a%5B9%5D = a%5B8%5D%2B7d,

    and then adding the terms.

    Hense, a%5B8%5D = 375%2F15 = 25.     (2)


3.  Thus from (1) we have a%5B6%5D = 38%2F2 = 19.   From (2),  a%5B8%5D = 25.

    From this, we immediately have 2d = a%5B8%5D-a%5B6%5D = 25-19 = 6.

    Hence, the common difference d = 6/2 = 3.


4.  Then a%5B1%5D = a%5B6%5D-5d = 19 - 5*3 = 19 - 15 = 4.


Answer.  a%5B1%5D = 4;  d = 3.

The lesson to learn from this solution: Sometimes, it is useful to think,
                    whether is it possible to build the solution around the properties of the central term of an arithmetic progression.


Problem 6

If  7  times the  7-th term of an AP is equal to  11  times its  11-th term,  show that its  18-th term is  0.

Solution 

We are given

7%2Aa%5B7%5D = 11%2Aa%5B11%5D.   ---->

7%2A%28a%5B1%5D%2B6d%29 = 11%2A%28a%5B1%5D%2B10%2Ad%29  ---->

%2811%2Aa%5B1%5D-7a%5B1%5D%29 + %28110d+-+42d%29 = 0  ---->

4a%5B1%5D + 68d = 0  --->  (cancel the factor 4 in both sides)  ---->

a%5B1%5D+%2B+17d = 0  --->

a%5B18%5D = 0.

QED.

Problem 7

The sum of the first five terms of an arithmetic progression is  10,  the sum of their squares is  380.
Find the first term and common difference.

Solution 

Let  x  be the  third  term of our arithmetic progression and  d  be its  common difference. 

Then the five first terms are


x - 2d
x - d
x
x + d
x + 2d.


The sum of these terms is  5x.  The sum of their squares is 5x%5E2 + 10d%5E2.  It is easy to check.


Thus we have the system of two  (non-linear)  equations


system%285x+=+10%2C%0D%0A5x%5E2+%2B+10d%5E2+=+380%29.


From the first equation we have  x = 10%2F5 = 2.

Substitute it into the second equation.  You will get 


5%2A2%5E2 + 10d%5E2 = 380,   or   10d%5E2 = 380+-+20 = 360.


Hence,  d%5E2 = 360%2F10 = 36  and  d = +/-6.


Therefore,  the members of the progression are 

-10, -4, 2, 8, 14


- in this order or in the opposite order - it doesn't matter.


You can check that this progression  (or these two progressions)  satisfies to the given conditions.

Problem 8

The coefficients  a,  b,  c of the quadratic equation  ax%5E2%2Bbx%2Bc = 0  form an arithmetic progression.  One root of this equation is  2.
Find the other root.

Solution 

You are given that

ax%5E2+%2B+bx+%2B+c = 0.                            (1)

Since a, b and c are AP, b = %28a%2Bc%29%2F2.       (2)  


    It is a characteristic property of any AP.
    See the lesson One characteristic property of arithmetic progressions in this site.


In (1), replace b by %28a%2Bc%29%2F2 and divide both sides of (1) by "a". You will get an equivalent equation

x%5E2+%2B+%281%2F2+%2B+c%2F%282a%29%29x+%2B+%28c%2Fa%29 = 0.                     (3)

Since x= 2 is the root of this equation (Given !), you can substitute x= 2 into (3). You will get

2%5E2+%2B+%281%2F2+%2B+c%2F%282a%29%29%2A2+%2B+%28c%2Fa%29 = 0,   or

4+%2B+1+%2B+c%2Fa+%2B+c%2Fa = 0,   or

2%2A%28c%2Fa%29 = -5   and finally  c%2Fa = -5%2F2.


Now your equation (3) takes the form

x%5E2+%2B+%281%2F2+-+5%2F4%29x+-+5%2F2 = 0,   or, which is the same,

x%5E2+-+%283%2F4%29x+-+5%2F2 = 0.                          (4)


    You may check that x= 2 is the root of this equation.

    By the way, you actually found THE COEFFICIENTS of your original equation (1)
       under the assumption that the leading coefficient "a" is equal to 1.
       These coefficients are 1, -3%2F4 and -5%2F2, and you can check that they form an AP.


Now you can easily find the other root from the equation (4) by using the quadratic formula.

It is -10%2F8 = -5%2F4.

Problem 9

The first three terms of an arithmetic progression are  x,  2x+1  and  5x+1.  Find  x  and the sum of the first  10  terms.

Solution 

a%5B2%5D - a%5B1%5D = a%5B3%5D - a%5B2%5D  ====>


(2x+1) -x = (5x+1) - (2x+1)    


x + 1 = 3x


1 = 2x  ====>  x = 1%2F2 is the first term.


(2x+1) - x = (1+1) - 1/2 = 3%2F2 is the common difference.


The 10-th term is   1%2F2 + 9%2A%283%2F2%29 = %281+%2B+27%29%2F2 = 14.


The sum of 10 terms = taken 10 times the average of the 1-st and the 10-th terms (which is equal to %2814%2B0.5%29%2F2 = 7.25): 


Sum = 7.25*10 = 72.5.


My other lessons on arithmetic progressions in this site are
    - Arithmetic progressions
    - 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 progressionss                                                                             (this lesson)
    - 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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