JEE Main201910 Jan 2019Evening ShiftMathematicsDeterminantsActual
Let a 1 , a 2 , a 3 … , a 10 be in G . P . with a i > 0 for i = 1 , 2 , … , 10 and S be the set of pairs r , k , r , k ∈ N (the set of natural numbers) for which log e ⁡ a 1 r a 2 k log e ⁡ a 2 r a 3 k log e ⁡ a 3 r a 4 k log e ⁡ a 4 r a 5 k log e ⁡ a 5 r a 6 k log e ⁡ a 6 r a 7 k log e ⁡ a 7 r a 8 k log e ⁡ a 8 r a 9 k log e ⁡ a 9 r a 10 k =
Options
- AInfinitely many
- B4
- C10
- D2
Correct answer
A. Infinitely many
Step-by-step solution
C 3 → C 3 - C 2 , C 2 → C 2 - C 1 ( Let α is common ratio of G P ) l o g e a 1 r a 2 k l o g e α r + k l o g e α r + k l o g e a 4 r a 5 k l o g e α r + k log e α r + k l o g e a 7 r a 8 k log e α r + k log e α r + k = 0 which is always true because C 2 & C 3 are identical.