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Let $l_{1}, l_{2}, ......., l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_{1}$ , and let $w_{1}, w_{2}, ........, w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_{2}$ , where $d_{1} d_{2} = 20 .$ For each $i = 1, 2, ........, 100$ , let $R_{i}$ be a rectangle with length $l_{i}$ , width $w_{i}$ and area $A_{i}$ . If $A_{41} - A_{31} = 14400$ , then the value of $\frac{A_{100} - A_{90}}{100}$ is _______________.

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answer is 38000.

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Detailed Solution

Given $A_{51} - A_{50} = 1000 \Rightarrow l_{51} w_{51} - l_{50} w_{50} = 1000$

$\Rightarrow (l_{1} + 40 d_{1}) (w_{1} + 40 d_{2}) - (l_{1} + 30 d_{1}) (w_{1} + 30 d_{2}) = 14400$

$(As d_{1} d_{2} = 20) \Rightarrow l_{1} d_{2} + w_{1} d_{1} = 40$

$∴ A_{100} - A_{90} = l_{100} w_{100} - l_{90} w_{90} = (l_{1} + 99 d_{1}) (w_{1} + 99 d_{2}) - (l_{1} + 89 d_{1}) (w_{1} + 89 d_{2})$

$= 10 (l_{1} d_{2} + w_{1} d_{1}) + (99^{2} - 89^{2}) d_{1} d_{2} = 10 (40) + \underset{= 10}{\underset{⎵}{(99 - 89)}} (99 + 89) (20)$

(As $d_{1} d_{2} = 20$ ) = 400 + 37600= 38000

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