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(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r} = 3$. Cubing both si

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$\frac{2}{3}$

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(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r} = 3$. Cubing both sides,we get $\frac{a^3}{(1-r)^3} = 27 \quad (1)$. The series of the cubes of the terms is $a^3, a^3r^3, a^3r^6, \dots$,which is also a geometric series with first term $a^3$ and common ratio $r^3$. The sum of this series is $\frac{a^3}{1-r^3} = \frac{27}{19} \quad (2)$. Dividing equation $(1)$ by equation $(2)$,we get $\frac{1-r^3}{(1-r)^3} = \frac{27}{27/19} = 19$. Using the identity $1-r^3 = (1-r)(1+r+r^2)$,we have $\frac{(1-r)(1+r+r^2)}{(1-r)^3} = 19$,which simplifies to $\frac{1+r+r^2}{(1-r)^2} = 19$. $1+r+r^2 = 19(1-2r+r^2) \implies 1+r+r^2 = 19-38r+19r^2$. $18r^2 - 39r + 18 = 0$. Dividing by $3$,we get $6r^2 - 13r + 6 = 0$. $(2r-3)(3r-2) = 0$. Since $|r| < 1$ for an infinite geometric series,$r = \frac{2}{3}$.

The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is

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