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(B) Let the infinite geometric series be $a, ar, ar^2, \dots$ where $a > 0$ and $|r| < 1$.The sum of the series is given by $S = \frac{a}{1-r} = 3 \im

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) Let the infinite geometric series be $a, ar, ar^2, \dots$ where $a > 0$ and $|r| The sum of the series is given by $S = \frac{a}{1-r} = 3 \implies a = 3(1-r)$.The cubes of the terms are $a^3, (ar)^3, (ar^2)^3, \dots$,which is a new geometric series with first term $a^3$ and common ratio $r^3$.The sum of the cubes is given by $S' = \frac{a^3}{1-r^3} = \frac{27}{19}$.Substituting $a = 3(1-r)$ into the second equation:$\frac{[3(1-r)]^3}{1-r^3} = \frac{27}{19}$$\frac{27(1-r)^3}{(1-r)(1+r+r^2)} = \frac{27}{19}$$\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$$19(1 - 2r + r^2) = 1 + r + r^2$$19 - 38r + 19r^2 = 1 + r + r^2$$18r^2 - 39r + 18 = 0$Dividing by $3$: $6r^2 - 13r + 6 = 0$$6r^2 - 9r - 4r + 6 = 0$$3r(2r - 3) - 2(2r - 3) = 0$$(3r - 2)(2r - 3) = 0$Since $|r| < 1$,we must have $r = \frac{2}{3}$.

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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