Question

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

• A. $ln8-ln3$
• B. $ln3-ln8$
• C. $ln24$
• D. $2ln6$

Answer 1

Answer: (A)

Condition $x^2>3$ for all logs to be defined
$2ln\left(x^2-1\right)=ln\left(2x^2-5\right)+ln\left(x^2-3\right)$
$\Rightarrow {\left(x^2-1\right)}^2=\left(2x^2-5\right)\left(x^2-3\right)$
$\Rightarrow x^4+1-2x^2=2x^4-6x^2-5x^2+15$
$\Rightarrow x^4-9x+14=0$ $\Rightarrow x^2=2,7$
Since, $x^2>3$ $\Rightarrow x^2=7$
So the numbers are $\to ln9,ln6,ln4$
$\Rightarrow d=ln6-ln9=ln\frac{2}{3}$
$\Rightarrow$ fourth term $=ln4+ln\frac{2}{3}=ln\frac{8}{3}$