Um problema de matemática do Egito Antigo era:
Dividir 100 pães entre 5 homens de tal forma que as porções recebidas devam estar em progressão aritmética crescente e que 1/7 da soma das três maiores porções seja igual à soma das duas menores. Qual será a porção do 1° homem?
Soluções para a tarefa
Respondido por
2
Boa tarde!
Solução!
Vamos organizar os dados do problema assim.
![a1=(x-2r)\\\\\
a2=(x-r)\\\\\\
a3=(x)\\\\\
a4=(x+r)\\\\\\
a5=(x+2r)\\\\\\
an=100](https://tex.z-dn.net/?f=a1%3D%28x-2r%29%5C%5C%5C%5C%5C%0Aa2%3D%28x-r%29%5C%5C%5C%5C%5C%5C%0Aa3%3D%28x%29%5C%5C%5C%5C%5C%0Aa4%3D%28x%2Br%29%5C%5C%5C%5C%5C%5C%0Aa5%3D%28x%2B2r%29%5C%5C%5C%5C%5C%5C%0Aan%3D100 "a1=(x-2r)\\\\\
a2=(x-r)\\\\\\
a3=(x)\\\\\
a4=(x+r)\\\\\\
a5=(x+2r)\\\\\\
an=100")
![(x-2r)+(x-r)+(x)+(x+r)+(x+2r)=100\\\\\\
5x-2r+2r-r+r=100\\\\\\
5x=100\\\\\\
x= \dfrac{100}{5}\\\\\\
\boxed{x=20}\\\\\](https://tex.z-dn.net/?f=%28x-2r%29%2B%28x-r%29%2B%28x%29%2B%28x%2Br%29%2B%28x%2B2r%29%3D100%5C%5C%5C%5C%5C%5C%0A5x-2r%2B2r-r%2Br%3D100%5C%5C%5C%5C%5C%5C%0A5x%3D100%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B100%7D%7B5%7D%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bx%3D20%7D%5C%5C%5C%5C%5C%0A+ "(x-2r)+(x-r)+(x)+(x+r)+(x+2r)=100\\\\\\
5x-2r+2r-r+r=100\\\\\\
5x=100\\\\\\
x= \dfrac{100}{5}\\\\\\
\boxed{x=20}\\\\\")
![x+(x+r)+(x+2r)=(x-2r)+(x-r)\\\\\
3x+3r=2x-3r\\\\\\
\dfrac{1}{7}(3x+3r)=2x-3r\\\\\\
\dfrac{3x+3r}{7}=2x-3r\\\\\\
3x+3r=14x-21r\\\\\\\\
3(20)+3r=14(20)-21r\\\\\\\
60+3r=280-21r\\\\\\\
3r+21r=280-60\\\\\\\
24r=220\\\\\\\
r= \dfrac{220}{24}\\\\\\\
Simplificando!\\\\\\\
r= \dfrac{55}{6}](https://tex.z-dn.net/?f=x%2B%28x%2Br%29%2B%28x%2B2r%29%3D%28x-2r%29%2B%28x-r%29%5C%5C%5C%5C%5C%0A3x%2B3r%3D2x-3r%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B1%7D%7B7%7D%283x%2B3r%29%3D2x-3r%5C%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B3x%2B3r%7D%7B7%7D%3D2x-3r%5C%5C%5C%5C%5C%5C%0A3x%2B3r%3D14x-21r%5C%5C%5C%5C%5C%5C%5C%5C%0A3%2820%29%2B3r%3D14%2820%29-21r%5C%5C%5C%5C%5C%5C%5C%0A60%2B3r%3D280-21r%5C%5C%5C%5C%5C%5C%5C%0A3r%2B21r%3D280-60%5C%5C%5C%5C%5C%5C%5C%0A24r%3D220%5C%5C%5C%5C%5C%5C%5C%0Ar%3D+%5Cdfrac%7B220%7D%7B24%7D%5C%5C%5C%5C%5C%5C%5C%0ASimplificando%21%5C%5C%5C%5C%5C%5C%5C%0Ar%3D+%5Cdfrac%7B55%7D%7B6%7D++ "x+(x+r)+(x+2r)=(x-2r)+(x-r)\\\\\
3x+3r=2x-3r\\\\\\
\dfrac{1}{7}(3x+3r)=2x-3r\\\\\\
\dfrac{3x+3r}{7}=2x-3r\\\\\\
3x+3r=14x-21r\\\\\\\\
3(20)+3r=14(20)-21r\\\\\\\
60+3r=280-21r\\\\\\\
3r+21r=280-60\\\\\\\
24r=220\\\\\\\
r= \dfrac{220}{24}\\\\\\\
Simplificando!\\\\\\\
r= \dfrac{55}{6}")
Como o primeiro homem esta representado pelo primeiro termo da P.A.
![x=20\\\\\\\
r= \frac{55}{6}\\\\\\\
a1=(x-2r)\\\\\\
a1=\bigg(20-2\times \dfrac{55}{6}\bigg)\\\\\\\
a1=\bigg(20 -\dfrac{110}{6}\bigg)\\\\\\\
a1=\bigg(\dfrac{120-110}{6}\bigg)\\\\\\\
a1= \dfrac{10}{6}\\\\\\
\boxed{a1= \dfrac{5}{3}}](https://tex.z-dn.net/?f=x%3D20%5C%5C%5C%5C%5C%5C%5C%0Ar%3D+%5Cfrac%7B55%7D%7B6%7D%5C%5C%5C%5C%5C%5C%5C%0Aa1%3D%28x-2r%29%5C%5C%5C%5C%5C%5C%0Aa1%3D%5Cbigg%2820-2%5Ctimes+%5Cdfrac%7B55%7D%7B6%7D%5Cbigg%29%5C%5C%5C%5C%5C%5C%5C%0A%0A++%0Aa1%3D%5Cbigg%2820+-%5Cdfrac%7B110%7D%7B6%7D%5Cbigg%29%5C%5C%5C%5C%5C%5C%5C%0A%0Aa1%3D%5Cbigg%28%5Cdfrac%7B120-110%7D%7B6%7D%5Cbigg%29%5C%5C%5C%5C%5C%5C%5C%0Aa1%3D+%5Cdfrac%7B10%7D%7B6%7D%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Ba1%3D+%5Cdfrac%7B5%7D%7B3%7D%7D++%0A++ "x=20\\\\\\\
r= \frac{55}{6}\\\\\\\
a1=(x-2r)\\\\\\
a1=\bigg(20-2\times \dfrac{55}{6}\bigg)\\\\\\\
a1=\bigg(20 -\dfrac{110}{6}\bigg)\\\\\\\
a1=\bigg(\dfrac{120-110}{6}\bigg)\\\\\\\
a1= \dfrac{10}{6}\\\\\\
\boxed{a1= \dfrac{5}{3}}")
Boa tarde!
Bons estudos!
Solução!
Vamos organizar os dados do problema assim.
Como o primeiro homem esta representado pelo primeiro termo da P.A.
Boa tarde!
Bons estudos!
Perguntas interessantes