Question

Let $l_1,l_2,\dots ,l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_1$, and let $w_1,w_2,\dots ,w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_2$, where $d_1{d}_2=10$. For each $i=1,2,\dots ,100$, let $R_l$ be a rectangle with length $l_i$, width $w_i$ and area $Ai$. If $A_{51}-A_{50}=1000$, then the value of $A_{100}-A_{90}$ is ___

Answer 1

Answer: 18900.00

Given
$A_{51}-A_{50}=1000\Rightarrow {\ell }_{51}{w}_{51}-{\ell }_{50}{w}_{50}=1000$
$\Rightarrow \left({\ell }_1+50d_1\right)\left(w_1+50d_2\right)-\left({\ell }_1+49d_1\right)\left(w_1+49d_2\right)=1000$
$\Rightarrow \left({\ell }_1{d}_2+w_1{d}_1\right)=10$
$\left(\text{ As }{d}_1{d}_2=10\right)$
$\therefore A_{100}-A_{90}={\ell }_{100}{w}_{100}-{\ell }_{90}{w}_{90}$
$=\left({\ell }_1+99d_1\right)\left(w_1+99d_2\right)-\left({\ell }_1+89d_1\right)\left(w_1+89d_2\right)$
$=10\left({\ell }_1{d}_2+w_1{d}_1\right)+\left(99^2-89^2\right)d_1{d}_2$
$=10(10)+\underset{=10}{\underbrace{(99-89)}}(99+89)(10)$
$\left(\text{ As, }{d}_1{d}_2=10\right)$
$=100(1+188)=100(189)$
$=18900$