Let $a_1, a_2,\dots ,a_{10}$ be in A.P and $h_1, h_2.....h_{10}$ be in H.P. If $a_1=h_1=2$ and

$a_{10}=h_{10}=3,$ then $a_4{h}_7$ is 

Let $d$ be the common difference of the A.P…., then 

$\begin{array}{l} 3=a_{10}=2+9d\Rightarrow d=1/9\\ \therefore a_1=2+3d=7/3\end{array}$

Next, let $D$ be the common difference of the A.P.

$\frac{1}{h_1},\frac{1}{h_2},\dots ,\frac{1}{h_{10}}$ Then 

$\begin{array}{l} \frac{1}{3}=\frac{1}{h_{10}}=\frac{1}{2}+9D\Rightarrow D=-\frac{1}{54}\\ \therefore \frac{1}{h_7}=\frac{1}{h_1}+6D=\frac{7}{18}\Rightarrow h_7=\frac{18}{7}\end{array}$

Hence, $a_4{h}_7=(7/3)(18/7)=6.$