notam distanta =x
avem distanta care mai este de parcurs inainte de ziua n: xₙ
[tex]x_n=16\times n+\frac{1}{3}(x_n-16\times n)[/tex]
[tex]x_5=16\times 5+\frac{1}{3}(x_5-16\times 5)\ \ \ |\times 5\\\\\\3x_5=240+x_5-80\\\\ 2x_5=160\\\\x_5=80\ km[/tex]
distanta parcursa in ziua 5=80 km
[tex]x_4-64-\frac{1}{3} (x_4-64)=80\ km[/tex]
[tex]3x_4-192-x_4+64=240\\\\2x_4=368\\\\x_4=184\ km\\[/tex]
distanta parcursa in ziua 4=184-80=104 km
[tex]x_3-48-\frac{1}{3}(x_3-48)=184\\\\ 2x_3-144+48=552\\\\x_3=648:2=324\ km[/tex]
distanta parcursa in ziua a treia=324-184=140 km
[tex]x_2-32-\frac{1}{3}(x-32)=324\\\\2x_2-96+32=972\\\\2x_2=1036\\\\x_2=518\ km[/tex]
distanta parcursa a doua zi=518-324=194 km
[tex]x_1-16-\frac{1}{3}(x-16)= 518\\\\2x_1-48+16=1554\\\\x_1=793\ km[/tex]
distanta parcursa in prima zi=793-518=275 km
Distanta traseu=275+194+140+104+80=793 km
