Question

Insert two harmonic means between 10 ,11.

• A. $\frac{110}{31},\frac{110}{30}$
• B. $\frac{330}{32},\frac{330}{31}$
• C. $\frac{32}{110},\frac{31}{110}$
• D. $31,32$

Answer 1

Answer: (B)

$\frac{330}{32},\frac{330}{31}$

$\Rightarrow$ Let $a=10,b=11$ be two given numbers and $H_{1}and{H}_2$ are $2$ harmonic mean between $a$ and $b$.

$\Rightarrow$ Now, $a,H_1,H_2,b$ are in $H.P.$

$\Rightarrow$ $\frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},\frac{1}{b}$ are in $A.P$.

$\Rightarrow$ Let $D$ be the common difference of this $A.P.$

$\Rightarrow$ $D=\frac{a-b}{(n+1)ab}=\frac{10-11}{(2+1)\times 10\times 11}=\frac{-1}{330}$

$\Rightarrow$ Now, $\frac{1}{H_1}=\frac{1}{a}+D=\frac{1}{10}+\left(\frac{-1}{330}\right)=\frac{32}{330}$ ...... $\left(1\right)$

$\Rightarrow$ $\frac{1}{H_2}=\frac{1}{a}+2D=\frac{1}{10}+2\times \left(\frac{-1}{330}\right)=\frac{31}{330}$ ....... $\left(2\right)$

Hence from $\left(1\right)$ and $\left(2\right)$, we get $H_1=\frac{330}{32},H_2=\frac{330}{31}$.