Question

fractia ordinara ireductibila rezultata din fractia 3+6+9+12+.....+150 supra 5+10+15+20+.....+250 prin simpkificarea acesteia este?

Answer 1

$\frac{3+6+9+...+150}{5+10+15+...+250}=\frac{3(1+2+3+...+50)}{5(1+2+3+...+50}=\frac{3}{5}$