Question

If a, b, c, are in H.P. then

• A. $\frac{b+a}{b-a}+\frac{b+c}{b-c}=2$
• B. $\frac{a-b}{b-c}=\frac{a}{c}$
• C. $\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}$
• D. All are correct

Answer 1

The correct option is D

All are correct

Since, a, b, c, are in H.P. therefore, $b=\frac{2ac}{a+c}...\left(i\right)$
Now $\frac{b}{a}=\frac{2c}{a+c}$
Applying componendo and dividendo, we get
$\frac{b+a}{b-a}=\frac{2c+a+c}{2c-a-c}=\frac{3c+a}{c-a}$
$Alsofrom\left(i\right)\frac{b}{c}=\frac{2a}{a+c}\Rightarrow \frac{b+c}{b-c}=\frac{3a+c}{a-c}\therefore \frac{b+a}{b-a}+\frac{b+c}{b-c}=\frac{3c+a}{c-a}+\frac{3a+c}{a-c}=\frac{3c+a-3a-c}{c-a}=\frac{2(c-a)}{c-a}=2$

Again since a,b,c, are in H.P.
$\therefore \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}\Rightarrow \frac{a-b}{ab}=\frac{b-c}{bc}\Rightarrow \frac{a-b}{b-c}=\frac{a}{c}$

$Next\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{\frac{2ac}{a+c}-a}+\frac{1}{\frac{2ac}{a+c}-c}=\frac{a+c}{ac-a^2}+\frac{a+c}{ac-c^2}=\frac{a+c}{a(c-a)}+\frac{a+c}{c(a-c)}=\frac{c+a}{c-a}[\frac{1}{a}-\frac{1}{c}]=\frac{c+a}{c-a}\left[\frac{c-a}{ca}\right]=\frac{c+a}{ca}=\frac{1}{c}+\frac{1}{a}$