Let the terms of infinite series are a,ar,ar2,ar3,… So, 1−ra=3 Since, sum of cubes of its terms is 1927 that is sum of a3, a3r3,…∞ is 1927 So, 1−r3a3=1927 ⇒1−ra×(1+r2+r)a2=1927 ⇒1+r2+r9(1+r2−2r)×3=1927 ⇒6r2−13r+6=0 ⇒(3r−2)(2r−3)=0 ⇒r=32, or 23 As ∣r∣<1 So, r=32
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…
JEE Main 2019 — Mathematics Algebra
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:
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
31
- B
32
- C
92
- D
94
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