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

Question

If ln ⁡ 2 x 2 - 5 , ln ⁡ x 2 - 1 and ln ⁡ x 2 - 3 are the first three terms of an arithmetic progression, then its fourth term is, ln ⁡ 8 - ln ⁡ 3, ln ⁡ 3 - ln 8, ln ⁡ 24, 2 ln ⁡ 6

MARKS App · Accepted Answer

Condition x 2 > 3 for all logs to be defined 2 ln ⁡ x 2 - 1 = ln ⁡ 2 x 2 - 5 + ln ⁡ x 2 - 3 ⇒ x 2 - 1 2 = 2 x 2 - 5 x 2 - 3 ⇒ x 4 + 1 - 2 x 2 = 2 x 4 - 6 x 2 - 5 x 2 + 1 5 ⇒ x 4 - 9 x + 14 = 0 ⇒ x 2 = 2,7 Since, x 2 > 3 ⇒ x 2 = 7 So the numbers are → ln ⁡ 9 , ln ⁡ 6 , ln ⁡ 4 ⇒ d = ln ⁡ 6 - ln ⁡ 9 = ln ⁡ 2 3 ⇒ fourth term = ln ⁡ 4 + ln ⁡ 2 3 = ln ⁡ 8 3

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