Let l_1, l_2, ldots, l_{100} be consecutive t | vedclass

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(A) Given $A_{51} - A_{50} = 1000$. Since $l_i = l_1 + (i-1)d_1$ and $w_i = w_1 + (i-1)d_2$,we have $A_i = l_i w_i = (l_1 + (i-1)d_1)(w_1 + (i-1)d_2)$

Vedclass · Accepted Answer

$18900$

Vedclass · Answer

(A) Given $A_{51} - A_{50} = 1000$. Since $l_i = l_1 + (i-1)d_1$ and $w_i = w_1 + (i-1)d_2$,we have $A_i = l_i w_i = (l_1 + (i-1)d_1)(w_1 + (i-1)d_2)$. $A_{51} - A_{50} = (l_1 + 50d_1)(w_1 + 50d_2) - (l_1 + 49d_1)(w_1 + 49d_2) = 1000$. Expanding this,we get $l_1 d_2 + w_1 d_1 + (50^2 - 49^2)d_1 d_2 = 1000$. Since $d_1 d_2 = 10$,this becomes $l_1 d_2 + w_1 d_1 + 99(10) = 1000$,which simplifies to $l_1 d_2 + w_1 d_1 = 10$. Now,$A_{100} - A_{90} = (l_1 + 99d_1)(w_1 + 99d_2) - (l_1 + 89d_1)(w_1 + 89d_2)$. $= (l_1 d_2 + w_1 d_1)(99 - 89) + (99^2 - 89^2)d_1 d_2$. $= 10(10) + (99 - 89)(99 + 89)(10) = 100 + 10(188)(10) = 100 + 18800 = 18900$.

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