+0  
 
0
50
3
avatar

Levans writes a positive fraction in which the numerator and denominator are integers, and the numerator is 1 greater than the denominator. He then writes several more fractions. To make each new fraction, he increases both the numerator and the denominator of the previous fraction by 1. He then multiplies all his fractions together. He has 20 fractions, and their product equals 3. What is the value of the first fraction he wrote?

 Nov 29, 2022
 #1
avatar+115 
-2

So first, lets understand this problem with an equation. This is what i got:

\(\frac{x}{x+1}\cdot\frac{x+1}{x+2} ...\frac{x+19}{x+20}=3\)

now we cancel. You see by cross canceling, the x+1's dissapear as all of them do and you are left with

\(\frac{x}{x+20} = 3\)

well, you might think that since it has to be a positive fraction and x has to be negative, this is wrong, BUT keep in mind, if both x in the numerator and denominator or negative, the fraction is a postive. so simplify this equation.

\(x = 3x+60\)  multiply both sides by \(x+20\)

\(2x=-60\)   subtracting x on both sides and moving the 60 over.

you will get x = -30, so the first fraction will be\(\frac{30}{31}\),which will be your answer!

 Nov 29, 2022
 #2
avatar
0

productfor(n, 1, 20, (10+n) / (9+n))==3

 

The first fraction ==11 / 10 and the last fraction ==30 / 29

 Nov 29, 2022
 #3
avatar
0

(x+1) / x  *  (x + 2) / (x +1) * (x + 3) / (x + 2) *...........* (x + 20) / (x + 19) = 3, solve for x

 

All cancel out except: (x + 20) / x = 3

 

x + 20 = 3x 

20 =3x - x = 2x

x = 20 / 2 = 10

 

Therefore, the first fraction ==10 + 1 / 10 = 11 / 10

 Nov 29, 2022

8 Online Users

avatar