Второй член геометрической прогрессии равен 9, а пятый член равен 1/3. Найдите члены прогрессии, заключенные между ними

[latex]b_2=9[/latex] [latex]b_5= \frac{1}{3} [/latex] [latex]b_3-[/latex] ? [latex]b_4-[/latex] ? [latex]b_n=b_1*q^{n-1}[/latex] [latex]b_5=b_1*q^4[/latex] [latex]b_2=b_1*q[/latex] [latex] \left \{ {{b_1*q=9} \atop {b_1*q^4= \frac{1}{3} }} \right. [/latex] [latex] \left \{ {{b_1*q=9} \atop {b_1*q*q^3= \frac{1}{3} }} \right. [/latex] [latex] \left \{ {{b_1*q=9} \atop {9*q^3= \frac{1}{3} }} \right. [/latex] [latex] \left \{ {{b_1*q=9} \atop {q^3= \frac{1}{27} }} \right. [/latex] [latex] \left \{ {{q= \frac{1}{3} } \atop {b_1* \frac{1}{3} =9}} \right. [/latex] [latex] \left \{ {{q= \frac{1}{3} } \atop {b_1} =27}} \right. [/latex] [latex]b_3=b_1*q^2=27*( \frac{1}{3} )^2=27*\frac{1}{9} =3[/latex] [latex]b_4=b_1*q^3=27*( \frac{1}{3} )^3=27*\frac{1}{27} =1[/latex] Ответ: 9; 3; 1; 1/3