Question 1028154: One day, Jack planted a magic beanstalk. (Let's call this day, "Day 1"). The next day (day 2), Jack notices that the beanstalk's height had increased by 1/2. On day 3, the beanstalk's height had increased by 1/3 of what it was on the previous day (day 2). One day 4, the beanstalk's height had increased by 1/4 of what it was on the previous day (day 3). Suppose the beanstalk continues to grow in this manner.
On what day will the beanstalk be 1000 times the height it was on the day Jack planted it?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! it appears that it is increasing by .5 each day.
assume day 1 is x
day 2 is x + 1/2 * x = 1.5 * x
day 3 is 1.5 * x + 1/3 * 1.5 * x = 2 * x
day 4 is 2 * x + 1/4 * 2 * x = 2.5 * x
day 5 is 2.5 * x + 1/5 * 2.5 * x = 3 * x
day 6 is 3 * x + 1/6 * 3 * x = 3.5 * x
day 7 is 3.5 * x + 1/7 * 3.5 * x = 4 * x
day 8 is 4 * x + 1/8 * 4 * x = 4.5 * x
etc.....
so .....
day 1 = x
day 2 = 1.5 * x
day 3 = 2 * x
day 4 = 2.5 * x
day 5 = 3 * x
day 6 = 3.5 * x
day 7 = 4 * x
etc....
the common difference appears to be .5 * x
let's assume that day 1 was 5 feet.
day 2 would be 5 + 1/2 * 5 = 7.5 feet.
day 3 would be 7.5 + 1/3 * 7.5 = 10 feet.
day 4 would be 10 + 1/4 * 10 = 12.5 feet.
day 5 would be 12.5 + 1/5 * 12.5 = 15 feet.
day 6 would be 15 + 1/6 * 15 = 17.5 feet.
day 7 would be 17.5 + 1/7 * 17.5 = 20 feet.
etc.....
the bean stalk is growing at 2.5 feet per day, assuming it's original height was 5 feet.
this make it an arithmetic progression because the common difference is 2.5 feet per day.
you want to know on what day it will be 1000 * it's original height.
the formula for the nth term of an arithmetic progression is An = A1 + (n-1)*d
An is the nth term in the progression.
A1 is the first term.
n is the number of terms
d is the common difference.
in our example above:
A1 = 5
An is equal to 1000 * 5 = 5000
n is what we want to find.
d is the common difference of 2.5
the formula becomes 5000 = 5 + (n-1) * 2.5
simplify to get 5000 = 5 + 2.5 * n - 2.5
combine like terms to get 5000 = 2.5 + 2.5 * n
subtract 2.5 from both sides of the equation to get 5000 - 2.5 = 2.5 * n
combine like terms to get 4997.5 = 2.5 * n
divide both sides of the equation by 2.5 to get 4997.5 / 2.5 = n
solve for n to get n = 4997.5 / 2.5 = 1999
it appears that the beanstalk will be 1000 times it's original height on day 1999.
let's see if we can get the same answer using x rather than 5.
A1 = x
An = 1000 * x
d = .5 * x
formula of An = A1 + (n-1) * d becomes (1000 * x) = x + (n-1) * (.5 * x)
simplify to get (1000 * x) = x + (.5 * x * n) - (1 * .5 * x)
simplify further to get (1000 * x) = x + (.5 * x * n) - (.5 * x)
combine like terms to get (1000 * x = (.5 * x) + (.5 * x * n)
subtract (.5 * x) from both sides of the equation to get (1000 * x) - (.5 * x) = (.5 * x * n)
simplify to get (999.5 * x) = (.5 * x * n)
divide both sides of the equation by (.5 * x) to get:
(999.5 * x) / (.5 * x) = n
the x in the numerator and denominator on the left side of the equation cancel out and you are left with:
999.5 / .5 = n
solve for n to get n = 999.5 / .5 = 1999.
you get that you get 1000 * the original value on day 1999.
let's see if that's true.
An = A1 + (n-1) * d
A1 = x
n = 1999
d = .5 * x
formula becomes An = x + 1998 * .5 * x
this results in An = x + 999 * x
combine like terms to get An = 1000 * x.
the solution is confirmed as good.
the key was to determine that it was an arithmetic progression.
this was determined by evaluating the difference between each successive term and finding that it was constant.
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