If a₁, a₂, a₃…. and b₁, b₂, b₃….. are A.P., and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 – a₁₀b₁₀, then a₄b₄ is equal to……...

$a_1,a_2,a_3......$ are in A.P. (Let common difference is d₁)

$b_1,b_2,b_3....$  are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$?a_1{b}_1=1$               

$\therefore b_1=\frac{1}{2}$

$a_{10}{b}_{10}=1$               

$\therefore b_{10}=\frac{1}{3}$                         

Now,

$a_{10}=a_1+9d_1\Rightarrow d_1=\frac{1}{9}$               

$b_{10}=b_1+9d_2\Rightarrow d_2=\frac{1}{9}\left[\frac{1}{3}-\frac{1}{2}\right]=-\frac{1}{54}$               

Now

$a_4=2+\frac{3}{9}=\frac{7}{3}$              

$b_4=\frac{1}{2}-\frac{3}{54}=\frac{4}{9}$             

$\therefore a_4{b}_4=\frac{28}{27}$