Question

If $a,b,c$are in $HP$, then which of the following is true.

• A. $\frac{1}{(b-a)}+\frac{1}{(b-c)}=\frac{1}{b}$
• B. $\frac{ac}{(a+c)}=b$
• C. $\frac{(b+a)}{(b-a)}+\frac{(b+c)}{(b-c)}=1$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Step 1. Given $a,b,c$ are in $H.P$

$\Rightarrow$ $b=2ac/(a+c)$

Step 2. Check Option $\left(1\right)$:

$[1/(b-a\left)\right]+[1/(b-c\left)\right]=1/b$

$\Rightarrow$$(b-c+b-a)/(b-a)(b-c)=1/b$

$\Rightarrow$ $(2b-c-a)/(b-a)(b-c)=1/b$

$\Rightarrow$ $b(2b-c-a)=(b-a)(b-c)$

$\Rightarrow$ $2b^2-bc-ab=b^2-ab-bc+ac$

$\Rightarrow$ $b^2=ac$

Option $\left(1\right)$ is wrong

Step 3. Check Option $\left(2\right)$;

$b=ac(a+c)$

Option $\left(2\right)$ was wrong

Step 4. Check Option $\left(3\right)$;

$\left[\right(b+a)/(b-a\left)\right]+\left[\right(b+c)/(b-c\left)\right]=1$

$\Rightarrow$$\left[\right(b+a\left)\right(b-c)+(b+c\left)\right(b-a\left)\right]/(b-a)(b-c)=1$

$\Rightarrow$ $(b+a)(b-c)+(b+c)(b-a)=(b-a)(b-c)$

$\Rightarrow$ $b^2-bc+ab-ac+b^2-ab+bc-ac=b^2-ab-bc+ac$

$\Rightarrow$ $b^2-2ac=-ab-bc+ac$

$\Rightarrow$ $b^2=-ab-bc+3ac$

Option $\left(3\right)$ was wrong

$\therefore$remaining Option $\left(4\right)$ is correct

Hence Option (4) is correct