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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 :
Option 1)

$\frac{4}{9}$
Option 2)

$\frac{1}{3}$
Option 3)

$\frac{2}{9}$
Option 4)

$\frac{2}{3}$

Answers (1)

best_answer

Sum of infinite terms of a GP -

$a+ar+ar^{2}+- - - - -= \frac{a}{1-r}\\here \left | r \right |<1$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

Lrt first term of G.P is a and the common ratio is r (r<1)

Given, $\frac{a}{1-r}=3$ 2) $a=3(1-r)---(1)$

Given, $\frac{a^{3}}{1-r^{3}}=\frac{27}{19}$ 2) $19a^{3}=27(1-r^{3})---(2)$

From (1) and (2)

$19[(1-r)]^{3}=27(1-r^{3})$

$19\times 27(1-r^{3})=27(1-r^{3})$

$19-38r+19r^{2}=1+r+r^{2}$

$18r^{2}-39r+18=0$

$\Rightarrow r=\frac{2}{3}$ since r<1

Option 1)

$\frac{4}{9}$

Option 2)

$\frac{1}{3}$

Option 3)

$\frac{2}{9}$

Option 4)

$\frac{2}{3}$

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