Q35 / 118 · NTA Abhyas (JEE Main) 2020 · Mathematics · Sequences and Series

If ln⁡2x2-5, ln⁡x2-1 and ln⁡x2-3 are the first three terms of an arithmetic progression, then its fourth term is

  • Correct answer
    A
    ln⁡8-ln⁡3
  • B
    ln⁡3-ln8
  • C
    ln⁡24
  • D
    2ln⁡6

Solution

Condition x2>3 for all logs to be defined
2ln⁡x2-1=ln⁡2x2-5+ln⁡x2-3
⇒x2-12=2x2-5x2-3
⇒x4+1-2x2=2x4-6x2-5x2+15
⇒x4-9x+14=0 ⇒x2=2,7
Since, x2>3 ⇒x2=7
So the numbers are →ln⁡9, ln⁡6, ln⁡4
⇒d=ln⁡6-ln⁡9=ln⁡23
⇒ fourth term =ln⁡4+ln⁡23=ln⁡83

Hence, the correct answer is option: 'A'.

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