QJEEAdv22Ma6886: JEE Advanced 2022 Mathematics - Class 11 JEEAdv22Ma6886 Solution: Let be consecutive terms of an arithmetic progression with…

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,

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

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

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

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

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

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

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

$= 18800 + 100$

$= 18900$