Question

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27/19. Then the common ratio of this series is:
(A) 1/3
(B) 2/3
(C) 2/9
(D) 4/9

Answer 1

HEY MATE YOUR ANSWER IS HERE....

YOUR ANSWER IS

OPTION ( B ) 2/3

EXPLAINATION :-

$\frac{a}{1 - r} = 3$

$a = 3(1 - r)$

$\frac{ {a}^{3} }{1 - {r}^{3} } = \frac{27}{19}$

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

$6 {r}^{2} - 13r + 6 = 0$

$r = \frac{2}{3} \: as \: |r| < 1$

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