Question

If $\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}$, then $a,b,c$ are in

• A. A.P
• B. G.P
• C. H.P
• D. In G.P and H.P both

Answer 1

The correct option is A A.P

$\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}\Rightarrow \frac{b-a+b-c}{(b-a)(b-c)}=\frac{c+a}{ac}\Rightarrow \frac{2b-a-c}{b^2+ac-bc-ba}=\frac{c+a}{ac}\Rightarrow 2abc-a^{2}c-a{c}^2=a{b}^2+c{a}^2-abc-{ba}^2+b^{2}c+a{c}^2-b{c}^2-abc\Rightarrow 2abc-a^{2}c-a{c}^2=-2abc+a{b}^2+c{a}^2-{ba}^2+b^{2}c+a{c}^2-b{c}^{2}4abc-a^{2}c-a{c}^2=a{b}^2+c{a}^2-{ba}^2+b^{2}c+a{c}^2-b{c}^{2}4abc=a{b}^2+c{a}^2-{ba}^2+b^{2}c+a{c}^2-b{c}^2+a^{2}c+a{c}^{2}4abc=2ac(a+c)+b^2(a+c)-b(a^2+c^2)$

by solving above equation we get

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

therefore $a,b,c$ are inA.P