Question

If $a,b,c$ are in H.P, then ${\log }_e(a+c)+{\log }_e(a-2b+c)$ is equal to ......................

• A. $2{\log }_e(c-a)$
• B. $2{\log }_e(b-a)$
• C. $2{\log }_e(b-c)$
• D. $2{\log }_e(a-c)$

Answer 1

The correct option is D $2{\log }_e(a-c)$

If $a,b,c$ are in H.P, then $\log (a+c)+\log (a-2b+c)$

$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$b=\frac{2ac}{a+c}$
$y=log(a+c)+log(a+c-2\frac{2ac}{a+c})$
$y=log(a+c)+log\left(\frac{a^2+2ac+c^2-4ac}{a+c}\right)$
$y=log(a+c)+2log(a-c)-log(a+c)$
$y=2log(a-c)$