SOLUTION: tim did 10 lunges on day 1 and continued this doing 4 more lunges on each day.on day 7th he took a breakand did nt do any day day.what is the equation for this? express it in the f
Algebra ->
Finance
-> SOLUTION: tim did 10 lunges on day 1 and continued this doing 4 more lunges on each day.on day 7th he took a breakand did nt do any day day.what is the equation for this? express it in the f
Log On
Question 1167337: tim did 10 lunges on day 1 and continued this doing 4 more lunges on each day.on day 7th he took a breakand did nt do any day 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 Answer by ikleyn(52884) (Show Source):
You can put this solution on YOUR website! 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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The numbers of lunges per day form arithmetic progression with the first term of = 10
and the common difference of d = 4.
Use the formula for the sum of first n terms of an arithmetic progression
= . (1)
In your problem
= .
They want you use x instead of n and y instead of .
OK. Then the formula takes the form
y = = 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
ANSWER. For the given sequence, the formula is y = 2x*(x+4).