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 :

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) (4/9) (B) (2/9) (C) (2/3) (D) (1/3)

Tardigrade Admin · Accepted Answer

(a/1-r) = 3 (a3/1-r3) = (27/19) ⇒ (27(1-r)3/1-r3) = (27/19) ⇒ 6r2 -13r+6 =0 ⇒ r = (2/3) as |r|<1

Q. 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 :

$\frac{4}{9}$

$\frac{2}{9}$

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{a}{1 - r} = 3$
$\frac{a ^{3}}{1 - r ^{3}} = \frac{27}{19} \Rightarrow \frac{27 ( 1 - r ) ^{3}}{1 - r ^{3}} = \frac{27}{19}$
$\Rightarrow 6 r^{2} - 13 r + 6 = 0$
$\Rightarrow r = \frac{2}{3}$ as $∣ r ∣ < 1$

Q. The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 1927​. Then the common ratio of this series is :

94​

92​

32​

31​