Sangeetha Pulapaka
2
1. A\ =\ 100\ \left(1\ -\ \frac{1}{2}\right)^{^n}
2. A\ =\ 200\ \left(1-\frac{1}{3}\right)^{^n}
3. No, because the initial amount is different, and the rate of decay is different. So the areas would never be the same, and the graphs would never meet at any point.

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Area of diagonal-generated triangles (video) | Khan Academy

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