a₁, a₂, a₃ .... are in A.P. (Let common difference is d₁)

b₁, b₂, b₃ .... are in A.P. (Let common difference is d₂)

and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$$\because$$ a₁b₁ = 1

$$\therefore$$ b₁ = $${1 \over 2}$$

a₁₀b₁₀ = 1

$$\therefore$$ b₁₀ = $${1 \over 3}$$

Now, a₁₀ = a₁ + 9d₁ $$\Rightarrow$$ d₁ = $${1 \over 9}$$

b₁₀ = b₁ + 9d₂ $$\Rightarrow$$ d₂ = $${1 \over 9}$$$$\left[ {{1 \over 3} - {1 \over 2}} \right]$$ = $$-$$ $${{1 \over {54}}}$$

Now, a₄ = 2 + $${{3 \over {9}}}$$ = $${{7 \over {3}}}$$

b₄ = $${{1 \over {2}}}$$ $$-%$$ $${{3 \over {54}}}$$ = $${{4 \over {9}}}$$

$$\therefore$$ a₄b₄ = $${{28 \over {27}}}$$