Question

If $ln \left(2 x^{2} - 5\right), \, ln⁡\left(x^{2} - 1\right)$ and $ln \left(x^{2} - 3\right)$ are the first three terms of an arithmetic progression, then its fourth term is

• A. $ln 8-ln ⁡ 3$
• B. $ln 3-ln8$
• C. $ln 24$
• D. $2ln 6$

Answer 1

Answer: (A)

Condition $x^{2}>3$ for all logs to be defined
$2ln \left(x^{2} - 1\right)=ln⁡\left(2 x^{2} - 5\right)+ln⁡\left(x^{2} - 3\right)$
$\Rightarrow \left(x^{2} - 1\right)^{2}=\left(2 x^{2} - 5\right)\left(x^{2} - 3\right)$
$\Rightarrow x^{4}+1-2x^{2}=2x^{4}-6x^{2}-5x^{2}+15$
$\Rightarrow x^{4}-9x+14=0$ $\Rightarrow x^{2}=2,7$
Since, $x^{2}>3$ $\Rightarrow x^{2}=7$
So the numbers are $ \rightarrow ln 9,ln⁡6,ln⁡4$
$\Rightarrow d=ln 6-ln ⁡ 9=ln ⁡ \frac{2}{3}$
$\Rightarrow $ fourth term $=ln 4+ln ⁡ \frac{2}{3}=ln ⁡ \frac{8}{3}$