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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 $R_i$ 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. . . . .
$18900$
$18901$
$18902$
$18903$
Solution
Given
$A _{51}- A _{50}=1000 \Rightarrow \ell_{S 1} w _{ S1 }-\ell_{50} w _{S 0}=1000$
$\Rightarrow\left(\ell_1+50 d _1\right)\left( w _1+50 d _2\right)-\left(\ell_1+49 d _1\right)\left( w _1+49 d _2\right)=1000$
$\Rightarrow\left(\ell_1 d _2+ w _1 d _1\right)=10$
(As $d _1 d _2=10$ )
$\therefore A _{100}- A _{90}=\ell_{100} w _{100}-\ell_{90} w _{90}$
$=\left(\ell_1+99 d _1\right)\left( w _1+99 d _2\right)-\left(\ell_1+89 d _1\right)\left( w _1+89 d _2\right)$
$=10\left(\ell_1 d _2+ w _1 d _1\right)+\left(99^2-89^2\right) d _1 d _2$
$=10(10)+\frac{(99-89)}{-10}(99+89)(10)$
$\left(\text { As, } d_1 d_2=10\right)$
$=100(1+188)=100(189)$
$=18900$
