If $a_1,a_2,a_3\dots$ and $b_1,b_2,b_3\dots .$ are A.P. and $a_1=2,a_{10}=3,a_1{b}_1=1=a_{10}{b}_{10}$ then $a_4{b}_4$ is equal to

$\begin{array}{r}{\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3\dots .\text{ A.P. };{\mathrm{a}}_1=2;{\mathrm{a}}_{10}=3;{\mathrm{d}}_1=\frac{1}{9}\\ {\mathrm{b}}_1,{\mathrm{b}}_2,{\mathrm{b}}_3,\dots \text{ A.P. };{\mathrm{b}}_1=\frac{1}{2};{\mathrm{b}}_{10}=\frac{1}{3};{\mathrm{d}}_2=\frac{-1}{54}\\ \left(\because {\mathrm{a}}_1{\mathrm{b}}_1=1={\mathrm{a}}_{10}{\mathrm{b}}_{10};{\mathrm{d}}_1\mathrm{\&}{\mathrm{d}}_2\right)\\ \begin{array}{l}{\mathrm{a}}_4\cdot {\mathrm{b}}_4=\left(2+3{\mathrm{d}}_1\right)\left(\frac{1}{2}+3{\mathrm{d}}_2\right)\\ =\left(2+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{18}\right)\\ =\left(\frac{7}{3}\right)\left(\frac{8}{18}\right)=\frac{28}{27}\end{array}\end{array}$