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8. Sequences and Series
hard
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
A
$\frac {1}{3}$
B
$\frac {2}{3}$
C
$\frac {2}{9}$
D
$\frac {4}{9}$
(JEE MAIN-2019)
Solution
$\frac{a}{{1 – r}} = 3$
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 $\frac{{1 – {r^3}}}{{{{(1 – r)}^3}}} = 19$
$r = \frac{2}{3}$
Standard 11
Mathematics
