Question 1167311
Tim did 10 lunges on day 1 and continued this doing 4 more lunges on each day. 
On day 7th he took a break and did not do any day. 
What is the equation for this? 
express it in the form y=ax(x+b)+c, where a, b and c are constants. x= no of days and y=no of lunges Tim Did
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<pre>
The numbers of lunges per day form arithmetic progression with the first term of  {{{a[1]}}} = 10
and the common difference of  d = 4.


Use the formula for the sum of first n terms of an arithmetic progression

    {{{S[n]}}} = {{{((2a[1] + (n-1)*d)*n)/2}}}.    (1)


In your problem

    {{{S[n]}}} = {{{((20 + (n-1)*4)*n)/2}}}.


They want you use x instead of n and y instead of {{{S[n]}}}.
OK.  Then the formula takes the form

    y = {{{((20 + 4(x-1))*x)/2}}} = 2(x-1)*x + 10x = 2x^2 - 2x + 10x = 2x^2 + 8x = 2x(x+4).


Thus, your final formula is

    y = 2x*(x+4),  1 <= x <= 6.    (2)


Comparing with the form  y = ax(x+b) + c,  you see for the coefficients 'a', 'b' and 'c'

    a = 2;  b = 4;  c = 0.         (3)


Below in the Table, calculations are compared using formulas (2), (3)  with direct calculations of the cumulative sum 


      T      A      B      L      E   


day   every   cumulative sum       cumulative sum
       day    computed directly    computed by formula (2)

 1     10         10                    10

 2     14         24                    24

 3     18         42                    42

 4     22         64                    64

 5     26         90                    90

 6     30        120                   120


<U>ANSWER</U>.  For the given sequence, the formula is  y = 2x*(x+4).
</pre>

Solved.