Question

If $a,b,c,ϵR^{+}$ = {$1$} and the numbers $lo{g}_{a}100,2lo{g}_{b}10,2lo{g}_{c}5+lo{g}_{e}4$ are in H.P then

• A. $2b=a+c$
• B. $b^2=ac$
• C. $b(a+c)=2ac$
• D. none of these

Answer 1

The correct option is B $b^2=ac$

$\frac{1}{{\log }_{a}100},\frac{1}{2{\log }_{b}10},\frac{1}{2{\log }_{c}5+{\log }_{c}4}$ are in $A\cdot P$

$\Rightarrow \frac{2}{2{\log }_{b}10}=\frac{1}{{\log }_{a}100}+\frac{1}{2{\log }_{c}5+2{\log }_{c}2}({\log }_{c}4=2{\log }_{2}2)$

$\Rightarrow \frac{1}{{\log }_{b}10}=\frac{1}{{\log }_{a}100}+\frac{1}{2{\log }_{c}10}({\log }_{c}5+{\log }_{c}2={\log }_c(5\times 2)={\log }_{c}10)$

$\Rightarrow {\log }_{10}b={\log }_{100}a+{\log }_{100}c(2{\log }_{c}10={\log }_c(10{)}^2)$

$\Rightarrow {\log }_{10}b={\log }_{100}\left(ac\right)$

$\Rightarrow {\log }_{10}b=\frac{1}{2}{\log }_{10}\left(ac\right)$

$\Rightarrow 2{\log }_{10}b={\log }_{10}\left(ac\right)$

$\Rightarrow {\log }_{10}{b}^2={\log }_{10}\left(ac\right)$

$\Rightarrow b^2=ac$

$\therefore$ option $B$ is correct.