Question

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?

Answer 1

Boa tarde!

Solução!

Vamos organizar os dados do problema assim.

$a1=(x-2r)\\\\\ a2=(x-r)\\\\\\ a3=(x)\\\\\ a4=(x+r)\\\\\\ a5=(x+2r)\\\\\\ an=100$

$(x-2r)+(x-r)+(x)+(x+r)+(x+2r)=100\\\\\\ 5x-2r+2r-r+r=100\\\\\\ 5x=100\\\\\\ x= \dfrac{100}{5}\\\\\\ \boxed{x=20}$


$x+(x+r)+(x+2r)=(x-2r)+(x-r)\\\\\ 3x+3r=2x-3r\\\\\\ \dfrac{1}{7}(3x+3r)=2x-3r\\\\\\ \dfrac{3x+3r}{7}=2x-3r\\\\\\ 3x+3r=14x-21r\\\\\\\\ 3(20)+3r=14(20)-21r\\\\\\\ 60+3r=280-21r\\\\\\\ 3r+21r=280-60\\\\\\\ 24r=220\\\\\\\ r= \dfrac{220}{24}\\\\\\\ Simplificando!\\\\\\\ r= \dfrac{55}{6}$

Como o primeiro homem esta representado pelo primeiro termo da P.A.


$x=20\\\\\\\ r= \frac{55}{6}\\\\\\\ a1=(x-2r)\\\\\\ a1=\bigg(20-2\times \dfrac{55}{6}\bigg)\\\\\\\ a1=\bigg(20 -\dfrac{110}{6}\bigg)\\\\\\\ a1=\bigg(\dfrac{120-110}{6}\bigg)\\\\\\\ a1= \dfrac{10}{6}\\\\\\ \boxed{a1= \dfrac{5}{3}}$

Boa tarde!
Bons estudos!