${{a}_{1}},{{a}_{2}},{{a}_{3}}..........{{a}_{10}}$ are in an arithmetic progression and ${{h}_{1}},{{h}_{2}},{{h}_{3}}..........{{h}_{10}}$ is in harmonic progression. The first terms of both the series are ${{a}_{1}}={{h}_{1}}=2$ and the last terms ${{a}_{10}}={{h}_{10}}=3$. Then find out the value of ${{a}_{4}}{{h}_{7}}$.
A. $2$
B. $3$
C. $5$
D. $6$

Hint: We will find the common difference for the first sequence by applying the formula for ${{n}^{th}}$ term that is ${{a}_{n}}=a+\left( n-1 \right)d$, where we will take $n=10$ , then after finding the common difference and then find the value of ${{a}_{4}}$ and then for the harmonic progression we will convert it into arithmetic progression and then again follow the same procedure and find the value of ${{h}_{7}}$ and then put those values into ${{a}_{4}}.{{h}_{7}}$ and get the answer.

Complete step-by-step answer:
We know that if a series is in arithmetic progression then the formula for ${{n}^{th}}$ term is as follows:
${{a}_{n}}=a+\left( n-1 \right)d$ ,
Where,
${{a}_{n}}={{n}^{th}}$ term of the A.P.
$a=$first term ,
$n=$ the total number of terms ,
$d=$ common difference
Now, we are given the first sequence as: ${{a}_{1}},{{a}_{2}},{{a}_{3}}..........{{a}_{10}}$ and also given that ${{a}_{1}}=2$ and ${{a}_{10}}=3$, so we will apply the formula for ${{n}^{th}}$ term, for $n=10$ :


Therefore,
$\begin{align}
  & \Rightarrow {{a}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow 3=2+\left( 10-1 \right)d \\
 & \Rightarrow 3-2=9d\Rightarrow d=\dfrac{1}{9} \\
\end{align}$

So, the common difference is: $d=\dfrac{1}{9}$
Now to find the value of ${{a}_{4}}$, we will apply the formula for ${{a}_{n}}$ for $n=4$ , $a=2$ and $d=\dfrac{1}{9}$ :
$\begin{align}
  & {{a}_{n}}=a+\left( n-1 \right)d\Rightarrow {{a}_{4}}=2+\left( 4-1 \right)\dfrac{1}{9} \\
 & {{a}_{4}}=2+\left( 3\times \dfrac{1}{9} \right)\Rightarrow {{a}_{4}}=2+\dfrac{1}{3} \\
 & {{a}_{4}}=\dfrac{7}{3}\text{ }................\text{ Equation 1} \\
\end{align}$

Now, it is given that ${{h}_{1}},{{h}_{2}},{{h}_{3}}..........{{h}_{10}}$ is in harmonic progression which means that \[\dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},\dfrac{1}{{{h}_{3}}},............\dfrac{1}{{{h}_{n}}}\] is in A.P.
We also know that the nth term of the Harmonic Progression (H.P.) :
$\dfrac{1}{{{h}_{n}}}=\dfrac{1}{a+\left( n-1 \right)d}$
Where,
${{h}_{n}}={{n}^{th}}$ term of the A.P.
$a=$first term ,
$n=$ the total number of terms ,
$d=$ common difference
To find the common difference we will apply this formula for ${{h}_{1}}=2$ and ${{h}_{10}}=3$, so we will apply the formula for ${{n}^{th}}$ term, for $n=10$ :
$\begin{align}
  & \Rightarrow \dfrac{1}{{{h}_{10}}}=\dfrac{1}{{{h}_{1}}}+\left( 10-1 \right)d\Rightarrow d=\dfrac{\dfrac{1}{{{h}_{10}}}-\dfrac{1}{{{h}_{1}}}}{9} \\
 & \Rightarrow d=\dfrac{\dfrac{1}{3}-\dfrac{1}{2}}{9}\Rightarrow d=\dfrac{\dfrac{2-3}{6}}{9}\Rightarrow d=\dfrac{\dfrac{-1}{6}}{9} \\
 & \Rightarrow d=\dfrac{-1}{54} \\
\end{align}$
Now to find the value of ${{h}_{7}}$, we will apply the formula for ${{a}_{n}}$ for $n=7$ , $a=\dfrac{1}{{{h}_{1}}}=\dfrac{1}{2}$ and $d=\dfrac{-1}{54}$ :
$\begin{align}
  & {{a}_{n}}=a+\left( n-1 \right)d\Rightarrow \dfrac{1}{{{h}_{7}}}=\dfrac{1}{{{h}_{1}}}+\left( 7-1 \right)\left( \dfrac{-1}{54} \right) \\
 & \dfrac{1}{{{h}_{7}}}=\dfrac{1}{2}-6\left( \dfrac{1}{54} \right)\Rightarrow \dfrac{1}{{{h}_{7}}}=\dfrac{7}{18} \\
 & {{h}_{7}}=\dfrac{18}{7}\text{ }...............\text{ Equation 2} \\
\end{align}$

Now to find out the value of ${{a}_{4}}{{h}_{7}}$ we will put the values from equation 1 and equation 2 :
$\begin{align}
  & {{a}_{4}}.{{h}_{7}}=\dfrac{7}{3}\times \dfrac{18}{7}=6 \\
 & {{a}_{4}}.{{h}_{7}}=6 \\
\end{align}$

So, the correct option is D.

Note: Students can make mistakes while applying the formula for ${{n}^{th}}$ term for the harmonic progression as the values are in harmonic progression. The formula is the same for the arithmetic progression and the sequence in harmonic progression as the reciprocals are in AP.