If a 1 , a 2 , a 3 … and b 1 , b 2 , b 3 … . are A.P. and a 1 = 2 , a 10 = 3 , a 1 b 1 = 1 = a…

Given, $a_1,a_2,a_3......\& b_1,b_2,b_3$.....are in $\mathrm{A}.\mathrm{P}$

Also given, $a_1=2, a_{10}=3$

and $a_1{b}_1=1=a_{10}{b}_{10}$

So $b_1=\frac{1}{a_1}=\frac{1}{2}$ and $b_{10}=\frac{1}{a_{10}}=\frac{1}{3}$

Now using $n^{th}$ term formula we get, $a_{10}=a_1+9d_a\Rightarrow 3=2+9d_a$

$\Rightarrow d_a=\frac{1}{9}$

Using same formula we get,

 $b_{10}=b_1+9d_b$  $\Rightarrow \frac{1}{3}=\frac{1}{2}+9d_b$  $\Rightarrow d_b=-\frac{1}{54}$

Now calculating $a_4{b}_4=\left(a_1+3d_a\right)\left(b_1+3d_b\right)$

$=\left(2+3\times \frac{1}{9}\right)\left(\frac{1}{2}+3\times \left(\frac{-1}{54}\right)\right)$

$= \frac{7}{3}\times \frac{8}{18}=\frac{56}{54}=\frac{28}{27}$