Given,

$${l_1},{l_2},\,.......,\,{l_{100}}$$ are in A.P with common difference $${d_1}$$.

So from property of A.P we can say,

$${l_2} = {l_1} + {d_1}$$

$${l_3} = {l_1} + 2{d_1}$$

$$ \vdots $$

$${l_{100}} = {l_1} + 99{d_1}$$

Also given,

$${w_1},{w_2},\,......\,,\,{w_{100}}$$ are in A.P with common difference $${d_2}$$.

$$\therefore$$ From the property of A.P we can say,

$${w_2} = {w_1} + {d_2}$$

$${w_3} = {w_1} + 2{d_2}$$

$$ \vdots $$

$${w_{100}} = {w_1} + 99{d_2}$$

Now, also given,

$${d_1}{d_2} = 10$$

and $${R_i}$$ is a rectangle whose length is $${l_i}$$ and width is $${w_i}$$ and area $${A_i}$$.

$$\therefore$$ We know, area of rectangle

$${A_i} = {l_i} \times {w_i}$$

$$\therefore$$ $${A_{51}} = {l_{51}} \times {w_{51}}$$

and $${A_{50}} = {l_{50}} \times {w_{50}}$$

Given, $${A_{51}} - {A_{50}} = 1000$$

$$ \Rightarrow ({l_1} + 50{d_1})({w_1} + 50{d_2}) - ({l_1} + 49{d_1})({w_1} + 49{d_2}) = 1000$$

$$ \Rightarrow [{l_1}{w_1} + 2500{d_1}{d_2} + 50({l_1}{d_2} + {d_1}{w_1})] - [{l_1}{w_1} + 49 \times 49{d_1}{d_2} + 49({l_1}{d_2} + {w_1}{d_1})] = 1000$$

$$ \Rightarrow [{(50)^2}{d_1}{d_2} - {(49)^2}{d_1}{d_2}] + (50 - 49)({l_1}{d_2} + {d_1}{w_1}) = 1000$$

$$ \Rightarrow (99 \times 1){d_1}{d_2} + {l_1}{d_2} + {d_1}{w_1} = 1000$$

$$ \Rightarrow 99 \times 10 + {l_1}{d_2} + {w_1}{d_1} = 1000$$

$$ \Rightarrow {l_1}{d_2} + {w_1}{d_1} = 10$$

Now,

$${A_{100}} - {A_{90}}$$

$$ = {l_{100}}\,.\,{w_{100}} - {l_{90}}\,.\,{w_{90}}$$

$$ = ({l_1} + 99{d_1})({w_1} + 99{d_2}) - ({l_1} + 89{d_1})({w_1} + 89{d_2})$$

$$ = [{l_1}{w_1} + {(99)^2}{d_1}{d_2} + 99({l_1}{d_2} + {w_1}{d_1})] - [{l_1}{w_1} + {(89)^2}{d_1}{d_2} + 89({l_1}{d_2} + {w_1}{d_1})]$$

$$ = [{(99)^2} - {89^2}]{d_1}{d_2} + 10({l_1}{d_2} + {w_1}{d_1})$$

$$ = 188 \times 10 \times {d_1}{d_2} + 10 \times 10$$

$$ = 188 \times 10 \times 10 + 100$$

$$ = 18800 + 100$$

$$ = 18900$$