Question

Insert three harmonic means between 2 and 3.

Answer 1

We solve this problem by starting with assuming the harmonic means between 2 and 3 as A, B, C. then we use the relation between A.P and H.P and say that $\dfrac{1}{2},\dfrac{1}{A},\dfrac{1}{B},\dfrac{1}{C},\dfrac{1}{3}$ are in A.P. Then we use the formula for ${{n}^{th}}$ term of A.P, ${{a}_{n}}=a+\left( n-1 \right)d$ and find the value of d by substituting $\dfrac{1}{3}$ as the fifth term. Then we use the same formula to find the values of A, B and C.

Complete step by step answer:
As we need to insert 3 harmonic means between 2 and 3, let us assume the harmonic means as A, B, C.
Then we can say that 2, A, B, C, 3 are in H.P, that is Harmonic Progression.
Now let us consider the property,
If a sequence of numbers ${{a}_{1}},{{a}_{2}},.....,{{a}_{n}}$ are in H.P then the sequence $\dfrac{1}{{{a}_{1}}},\dfrac{1}{{{a}_{2}}},\dfrac{1}{{{a}_{3}}},.......,\dfrac{1}{{{a}_{n}}}$ are in A.P.
So, by using this property, we get that $\dfrac{1}{2},\dfrac{1}{A},\dfrac{1}{B},\dfrac{1}{C},\dfrac{1}{3}$ are in A.P. So, the terms are
$\begin{aligned} & \Rightarrow {{a}_{1}}=\dfrac{1}{2} \\\ & \Rightarrow {{a}_{2}}=\dfrac{1}{A} \\\ & \Rightarrow {{a}_{3}}=\dfrac{1}{B} \\\ & \Rightarrow {{a}_{4}}=\dfrac{1}{C} \\\ & \Rightarrow {{a}_{5}}=\dfrac{1}{3} \\\ \end{aligned}$
As they are in A.P and we can see that the first term is $\dfrac{1}{2}$, let us assume that the common difference is $d$.
So, now let us consider the formula for the ${{n}^{th}}$ term of A.P with the first term a and common difference d,
${{a}_{n}}=a+\left( n-1 \right)d$
As we have that $\dfrac{1}{3}$ is the fifth term, using the above formula we have,
$\begin{aligned} & \Rightarrow \dfrac{1}{2}+\left( 5-1 \right)d=\dfrac{1}{3} \\\ & \Rightarrow 4d=\dfrac{1}{3}-\dfrac{1}{2} \\\ & \Rightarrow 4d=\dfrac{-1}{6} \\\ & \Rightarrow d=\dfrac{-1}{24} \\\ \end{aligned}$
Now as we have the value of $d$, we can find the second, third and fourth terms using the formula for the ${{n}^{th}}$ term of A.P and thereby find the values of A.
$\begin{aligned} & \Rightarrow \dfrac{1}{A}=\dfrac{1}{2}+\left( 2-1 \right)\left( \dfrac{-1}{24} \right) \\\ & \Rightarrow \dfrac{1}{A}=\dfrac{1}{2}+\dfrac{-1}{24} \\\ & \Rightarrow \dfrac{1}{A}=\dfrac{11}{24} \\\ & \Rightarrow A=\dfrac{24}{11}...............\left( 1 \right) \\\ \end{aligned}$
Now consider the third term,
$\begin{aligned} & \Rightarrow \dfrac{1}{B}=\dfrac{1}{2}+\left( 3-1 \right)\left( \dfrac{-1}{24} \right) \\\ & \Rightarrow \dfrac{1}{B}=\dfrac{1}{2}+\left( 2\times \dfrac{-1}{24} \right) \\\ & \Rightarrow \dfrac{1}{B}=\dfrac{1}{2}-\dfrac{1}{12} \\\ & \Rightarrow \dfrac{1}{B}=\dfrac{5}{12} \\\ & \Rightarrow B=\dfrac{12}{5}...............\left( 2 \right) \\\ \end{aligned}$
Now, consider the fourth term,
$\begin{aligned} & \Rightarrow \dfrac{1}{C}=\dfrac{1}{2}+\left( 4-1 \right)\left( \dfrac{-1}{24} \right) \\\ & \Rightarrow \dfrac{1}{C}=\dfrac{1}{2}+\left( 3\times \dfrac{-1}{24} \right) \\\ & \Rightarrow \dfrac{1}{C}=\dfrac{1}{2}-\dfrac{1}{8} \\\ & \Rightarrow \dfrac{1}{C}=\dfrac{3}{8} \\\ & \Rightarrow C=\dfrac{8}{3}...............\left( 3 \right) \\\ \end{aligned}$

So, we get the values of A, B, C as $\dfrac{24}{11},\dfrac{12}{5},\dfrac{8}{3}$.
So, the three harmonic means between 2 and 3 are $\dfrac{24}{11},\dfrac{12}{5},\dfrac{8}{3}$.
Hence answer is $\dfrac{24}{11},\dfrac{12}{5},\dfrac{8}{3}$.

Note:
We can also solve this question in an alternative easier method.
First let us consider the formula,
Harmonic mean of two numbers a and b is $\dfrac{2ab}{a+b}$
So, first let us insert 1 harmonic mean between them, that is
$\Rightarrow \dfrac{2\left( 2 \right)\left( 3 \right)}{2+3}=\dfrac{12}{5}$
So, our sequence becomes $2,\dfrac{12}{5},3$.
Now, let us find the harmonic mean between 2 and $\dfrac{12}{5}$.
$\Rightarrow \dfrac{2\left( 2 \right)\left( \dfrac{12}{5} \right)}{2+\dfrac{12}{5}}=\dfrac{\dfrac{48}{5}}{\dfrac{22}{5}}=\dfrac{48}{22}=\dfrac{24}{11}$
Now let us find the harmonic mean between $\dfrac{12}{5}$ and 3.
$\Rightarrow \dfrac{2\left( \dfrac{12}{5} \right)\left( 3 \right)}{\dfrac{12}{5}+3}=\dfrac{\dfrac{72}{5}}{\dfrac{27}{5}}=\dfrac{72}{27}=\dfrac{8}{3}$
So, we the sequence as $2,\dfrac{24}{11},\dfrac{12}{5},\dfrac{8}{3},3$.
Hence answer is $\dfrac{24}{11},\dfrac{12}{5},\dfrac{8}{3}$.