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Найти сумму членов бесконечно убывающей геометрической прогрессии если третий член этой прогрессии равен 2, а шестой равен 1/4

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Эксперт5

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$b_3=2\; \; \; b_6= \frac{1}{4}\; \; \; S-?\\\\ \left \{ {{b_3=2} \atop {b_6= \frac{1}{4} } \right.\\ \left \{ {{b_1*q^2=2} \atop {b_1*q^5= \frac{1}{4} }} \right. \\ \left \{ {{b_1= \frac{2}{q^2} } \atop { \frac{2}{q^2}*q^5= \frac{1}{4}}} \right. \\ \left \{ {{b_1= \frac{2}{q^2} } \atop {2q^3= \frac{1}{4} }} \right.\\ \left \{ {{b_1= \frac{2}{q^2} } \atop {q^3= \frac{1}{8} }} \right. \\ \left \{ {{b_1= \frac{2}{q^2} } \atop {q= \frac{1}{2} }} \right.\\$
$\left \{ {{b_1= \frac{2}{(1/2)^2} } \atop {q=1/2}} \right. \\ \left \{ {{b_1= \frac{2}{1/4} } \atop {q=1/2}} \right. \\ \left \{ {{b_1=8} \atop {q=1/2}} \right. \\\\S= \frac{b_1}{1-q}= \frac{8}{1- \frac{1}{2} }= \frac{8}{1/2}=8*2=16$

Ответ: 16

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