If a _ 1 , a _ 2 , a _ 3 and b _ 1 , b _ 2 , b _ 3 . are A.P. and a...

Given, ${a}_{1}, {a}_{2}, {a}_{3} . . . . . . \text{ and } {b}_{1}, {b}_{2}, {b}_{3}$.....are in $A . P$

Also given, ${a}_{1} = 2, {a}_{10} = 3$

and ${a}_{1}{b}_{1} = 1 = {a}_{10}{b}_{10}$

So ${b}_{1} = \dfrac{1}{{a}_{1}} = \dfrac{1}{2}$ and ${b}_{10} = \dfrac{1}{{a}_{10}} = \dfrac{1}{3}$

Now using ${n}^{th}$ term formula we get, ${a}_{10} = {a}_{1} + 9{d}_{a} ⇒ 3 = 2 + 9{d}_{a}$

$⇒ {d}_{a} = \dfrac{1}{9}$

Using same formula we get,

${b}_{10} = {b}_{1} + 9{d}_{b}$ $⇒ \dfrac{1}{3} = \dfrac{1}{2} + 9{d}_{b}$ $⇒ {d}_{b} = - \dfrac{1}{54}$

Now calculating ${a}_{4}{b}_{4} = \left({a}_{1} + 3{d}_{a}\right)\left({b}_{1} + 3{d}_{b}\right)$

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

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