Question

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

Answer 1

For the arithmetic progressions $l_1, l_2, \ldots, l_{100} \quad \text{and} \quad w_1, w_2, \ldots, w_{100}$

Let $T_1​=a$ and the common difference be $d_1$​, and similarly for $w_1, w_2, \ldots, w_{100}$

Let $T_1​=b$ and the common difference be $d_2​.$

Then, $A_{51} - A_{50} = l_{51}w_{51} - l_{50}w_{50}$

$(a+50d_1)(b+50d_2) - (a+49d_1)(b+49d_2)$

$(50bd_1+50ad_2+2500d_1d_2) - (49ad_2+49bd_1+2401d_1d_2)$

$bd_1 + ad_2 + 99d_1d_2 = 1000$

Therefore, $bd_1 + ad_2 = 10 \quad \text{(as } d_1d_2 = 10\text{)}$ denoted as equation $(i).$

Also, $A_{100} - A_{90} = l_{100}w_{100} - l_{90}w_{90}$

$=(a+99d1​)(b+99d2​)−(a+89d1​)(b+89d2​),$

$(99bd_1+99ad_2+992d_1d_2) - (89bd_1+89ad_2+892d_1d_2)$

$10(bd_1+ad_2)+1880d_1d_2$

$=10(10)+18800,$

$=18900.$