Lesson Entertainment problems on arithmetic progressions

Entertainment problems on arithmetic progressions

Problem 1

The numerator is


    8 + 16 + 24 + . . . + 784 = 8*(1 + 2 + 3 + . . . 98).


The denominator is


    6 + 12 + 18 + . . . + 588 = 6*(1 + 2 + 3 + . . . 98).


When you take the fraction, the sums in the parentheses cancel each other - - - so THERE IS NO NEED to calculate them.


The final answer is   = .

Problem 2

Without computing, it is clear that the number of addends is the same in both sums.


From the other side, each addend in the first sum is 1 unit greater than the corresponding addend in the second sum.


It leads us to the conclusion that the O-value is greater than E-value.



Next, O-value is greater than E-value exactly by the number of addends in each sum.


The number of intervals of the length 2 between  4 and 104 inclusive is   =  = 50.


Hence the number of terms in each sum is 50+1 = 51.


Thus O-value is 51 units greater than E-value.

Problem 3

There are 5700*66 portions of provision in the storage.


In the 1st day, 5700 portions are consumed.

In the 2nd day, 5700-20 = 5680 portions are consumed.

In the 3nd day, 5680-20 = 5660 portions are consumed.

. . . and so on . . . 


The problem wants you find the sum of the first n terms of this arithmetic progression

     = 


and the number n in such a way that  = 5700*66, under an additional condition  >= 0.


The first term of this AP is 5700; the common difference is -20,  so the sum of the first n terms is

     =  =  = .


So we write this equation

     = 5700*66


Simplify and find n

    20n^2 - 11420n + 11400*66 = 0

      n^2 - 571 + 5700*66 = 0.


Next, apply the quadratic formula.  You will get

       = .


The value  =  = 76 works: it satisfies  =  = 5700*66 and  >= 0.


The other value  =  = 495 does not work:  = 5700-20*495 = -4200 is negative.


So, the problem is just solved, and the ANSWER is 76 days.

Problem 4

Let x be the arithmetic mean of that (first mentioned) p consecutive integers.

Then the sum of that (first mentioned) p consecutive integers is x*p.


The average of the next p consecutive integers is  (x+p),

and the sum of the next p consecutive integers is (x+p)*p.


So, we have this equation

    (x+p)*p = x*p + 400

     xp + p^2 = xp + 400

          p^2 = 400

          p   = 20.


ANSWER.  p = 20.

Problem 5

You are given first three terms of an arithmetic progression 7, 5, 3.

As you see, each next term is 2 units less than the previous term.


Using this rule, continue the sequence further

    7, 5, 3, 1, -1, -3, -5, -7, -9.


Now take the sum of written terms.


Do not worry: it is easy to calculate the sum, because in the sum 
the negative terms will kill the positive terms.


The only term, which will survive is the term -9.


So, the sum of this sequence is -9.


At the same time, the term -9 is the 9-th term of the progression.


Doing this way, you just answered the problem's question:


    +---------------------- A N S W E R --------------------------+
    |      the value of "n" (the number of terms) when            |
    |    the sum of progression is equal to its n-th term, is 9.  |
    +-------------------------------------------------------------+

Problem 6

The given parameters are:

    Inner diameter  =  8 cm

    Outer diameter  = 60 cm 


The area between the two circles is

    Area =  -  =  = 2777.167903 cm^2.


The length of the paper is then calculated as:

    Length =  centimeters.


This gives

   Length =  = 555433.5806 cm,  or  5554.34 meters.  (rounded).  


ANSWER.  The length of the paper in the roll is 5554.34 meters or 5.55434 kilometer (rounded).

Problem 7

The minimum diameter (the diameter of the first layer) is d = 3.6 cm; 
the circumference (the length) of the first layer is   = 3.14159*3.6 = 11.309724 cm.


The maximum diameter (the diameter of the last layer) is D = 5.4 cm; 
the circumference (the length) of the last layer is   = 3.14159*5.4 = 16.964586 cm.


The lengths of consecutive layers make an arithmetic progression with the common  difference  ,
where    is the thickness of the tape. Therefore, the number of layers n can be found from 
the formula of the sum of an arithmetic progression


    2540 =  =  = 14.137155*n,


    n =  = 179.67  (approximately).


ANSWER.  The number of layers on the tape is about 180.