If the harmonic mean between a and b be H, then the value of (1/H-a)+(1/H-b) is

Question

If the harmonic mean between a and b be H, then the value of (1/H-a)+(1/H-b) is (A) a+b (B) a b (C) (1/a)+(1/b) (D) (1/a)-(1/b)

Tardigrade Admin · Accepted Answer

Putting H=(2 a b/a+b),, we have (1/H-a)+(1/H-b) =(1/((2 a b)a+b-a))+(1/((2 a b)a+b-b))=(a+b/a b-a2)+(a+b/a b-b2) =((a+b/b-a))((1/a)-(1/b)) =((a+b/b-a))((b-a/a b)) =(a+b/a b)=(1/a)+(1/b)

Q. If the harmonic mean between $a$ and $b$ be $H$ , then the value of $\frac{1}{H - a} + \frac{1}{H - b}$ is

$a + b$

$ab$

$\frac{1}{a} + \frac{1}{b}$

$\frac{1}{a} - \frac{1}{b}$

Q. If the harmonic mean between a and b be H, then the value of H−a1​+H−b1​ is

a+b

ab

a1​+b1​

a1​−b1​

Putting H=a+b2ab​,, we have H−a1​+H−b1​ =(a+b2ab​−a)1​+(a+b2ab​−b)1​=ab−a2a+b​+ab−b2a+b​ =(b−aa+b​)(a1​−b1​) =(b−aa+b​)(abb−a​) =aba+b​=a1​+b1​

Q. If the harmonic mean between $a$ and $b$ be $H$ , then the value of $\frac{1}{H - a} + \frac{1}{H - b}$ is

$a + b$

$ab$

$\frac{1}{a} + \frac{1}{b}$

$\frac{1}{a} - \frac{1}{b}$