The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 2719 . Then the common ratio of this series is:

Question

Accepted Answer

Correct answer: B. (b) JEE Main 2019 Let the terms of infinite series are a , ar ,ar^2 ,ar^3 , … So, a1-r=3 Since, sum of cubes of its terms is 2719 that is sum of a^3 ,a^3r^3 , … is 2719 So, a^31-r^3= 2719 ⇒ a1-r a^2(1+r^2+r)= 2719 ⇒ 9(1+r^2-2r) 31+r^2+r= 2719 ⇒6r^2-13r+6=0 ⇒ (3r-2 ) (2r-3 )=0 ⇒r= 23 , or 23 As |r |<1 So, r= 23

$The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is \frac{27}{19} . Then the common ratio of this series is:$

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