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 = 10 . For each

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 $R_{i}$ 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 _______. [2022]

Smart Achievers · Accepted Answer

(18900)

For A.P. $l_{1}, l_{2}, \dots, l_{100}$

$Let T_{1} = a and common difference = d_{1}$ $and similarly now for A.P. w_{1}, w_{2}, \dots, w_{100}$

$let T_{1} = b and common difference = d_{2}$

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

$= (a + 50 d_{1}) (b + 50 d_{2}) - (a + 49 d_{1}) (b + 49 d_{2})$

$= 50 b d_{1} + 50 a d_{2} + 2500 d_{1} d_{2} - 49 a d_{2} - 49 b d_{1} - 2401 d_{1} d_{2}$

$= b d_{1} + a d_{2} + 99 d_{1} d_{2} = 1000$

$\Rightarrow b d_{1} + a d_{2} = 1000 - 990 = 10 ...(i) (As d_{1} d_{2} = 10)$

$∴$ $A_{100} - A_{90} = l_{100} w_{100} - l_{90} w_{90}$

$= (a + 99 d_{1}) (b + 99 d_{2}) - (a + 89 d_{1}) (b + 89 d_{2})$

$= 99 b d_{1} + 99 a d_{2} + 99^{2} d_{1} d_{2} - 89 b d_{1} - 89 a d_{2} - 89^{2} d_{1} d_{2}$

$= 10 (b d_{1} + a d_{2}) + 1880 d_{1} d_{2}$

$\Rightarrow 10 (10) + 1880 (10) = 18900$

Solution

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