Răspuns:
Explicație pas cu pas:
[tex]b_{3}=8,~9*(b_{1}+b_{2}+b_{3})=b_{1}+b_{2}+b_{3}+b_{4}+b_{5}+b_{6},\\9*(b_{1}+b_{2}+b_{3})-(b_{1}+b_{2}+b_{3})=b_{4}+b_{5}+b_{6},\\8*(b_{1}+b_{2}+b_{3})=b_{4}+b_{5}+b_{6},~dar~b_{4}=b_{1}*q^{3},~b_{5}=b_{2}*q^{3},~b_{6}=b_{3}*q^{3},\\Atunci~8*(b_{1}+b_{2}+b_{3})=q^{3}*(b_{1}+b_{2}+b_{3}).~Deci~8=q^{3},~deci~q=2.\\Din~b_{3}=8,~b_{1}*q^{2}=8,~deci~b_{1}*4=8.~Deci~b_{1}=2.[/tex]
Cele 6 numere cautate sunt: 2,4,8,16,32,64.