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

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

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

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