JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 16)
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $${{27} \over {19}}$$.Then the common ratio of this series is :
$${4 \over 9}$$
$${1 \over 3}$$
$${2 \over 3}$$
$${2 \over 9}$$
Explanation
$${a \over {1 - r}} = 3\,\,\,\,\,\,\,.....(1)$$
$${{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}$$
$$ \Rightarrow 6{r^2} - 13r + 6 = 0$$
$$ \Rightarrow r = {2 \over 3}\,\,$$
as $$\left| r \right| < 1$$
$${{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}$$
$$ \Rightarrow 6{r^2} - 13r + 6 = 0$$
$$ \Rightarrow r = {2 \over 3}\,\,$$
as $$\left| r \right| < 1$$
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