Let  $l_1,l_2,\mathrm{......}{l}_{100}$ be consecutive terms of an A.P with common difference  $d_1$  and let  $w_1,w_2,\mathrm{....}{w}_{100}$  be the consecutive terms of another arthmatics progression with common difference  $d_2$  where  $d_1{d}_2=10$  for each  $i=1,2,\mathrm{...},100$  let  $R_1$  be a rectangle with length  $L_i$, width $W_i$  and area  $A_i$. If  $A_{51}-A_{50}=1000$  value of  $A_{100}-A_{90}$  is 

For A.P  $l_1,l_2,\mathrm{......}{l}_{100}$
Let $T_1=a$   and common difference  $=d_1$  and similarly for A.P  $w_1,w_2,\mathrm{....}{w}_{100}$
$T_1=b$  and common difference  $=d_2$
 $$\begin{gathered} A_{51}-A_{50}=l_{51}{w}_{51}-l_{50}{w}_{50} \\ =(a+50d_1)(b+50d_2)-(a+49d_1)(b+49d_2) \\ =50b{d}_1+50a{d}_2+2500d_1{d}_2-49a{d}_2-49b{d}_1-2401d_1{d}_2 \\ \Rightarrow \therefore b{d}_1+a{d}_2=\mathrm{10......}\left(1\right)(As{d}_1{d}_2=10) \\ \therefore A_{100}-A_{90}=l_{100}{w}_{100}-l_{90}{w}_{90} \\ =(a+99d_1)(b+99d_2)-(a+89d_1)(b+89d_2) \\ =99b{d}_1+99a{d}_2+{99}^2{d}_1{d}_2-89b{d}_1-89a{d}_2-{89}^2{d}_1{d}_2 \\ \begin{array}{l}\Rightarrow 10(b{d}_1+a{d}_2)+1800d_1{d}_2\\ \Rightarrow 10\left(10\right)+1800\Rightarrow 18,900\end{array} \end{gathered}$$