The sum of an infinite geometric series with | vedclass

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Question

$\frac{a}{{1 - r}} = 3$

Cube both sides

$\frac{{{a^3}}}{{{{(1 - r)}^3}}} = 27\,\,\,\,......\left( 1 \right)$

and $\frac{{{a^3}}}{{1 - {r

Vedclass · Accepted Answer

$\frac {2}{3}$

Vedclass · Answer

$\frac{a}{{1 - r}} = 3$

Cube both sides

$\frac{{{a^3}}}{{{{(1 - r)}^3}}} = 27\,\,\,\,......\left( 1 \right)$

and $\frac{{{a^3}}}{{1 - {r^3}}} = \frac{{27}}{{19}}\,\,\,\,......\left( 2 \right)$

$(1)/(2)$ given $\frac{{1 - {r^3}}}{{{{(1 - r)}^3}}} = 19$

$r = \frac{2}{3}$

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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