💡First, I recognize that Darcie is 4 years old and that she is one-sixth as old as her mother. This means I can set up an equation to find her mother's age.Next, I use the information that Darcie's mother is four-fifths as old as her father to establish another equation. By solving these equations step by step, I can determine the father's age.Finally, by performing the calculations, I find out that Darcie's father is 30 years old.

💡Alright, so I've got this geometry problem here, and I need to figure out under which conditions the proposition "If x is perpendicular to y, and y is parallel to z, then x is perpendicular to z" holds true. There are four different conditions given, each specifying whether x, y, and z are lines or planes. Let me go through each condition one by one and see if the proposition is valid in those cases.First, let's understand the proposition itself. It's saying that if x is perpendicular to y, and y is parallel to z, then x must be perpendicular to z. So, I need to see in which of the four scenarios this implication always holds true.Condition ①: x, y, z are all lines.Okay, so if x, y, and z are all lines, and x is perpendicular to y, and y is parallel to z, does that mean x is perpendicular to z? Let me visualize this. If y is parallel to z, then they have the same slope or direction. If x is perpendicular to y, then it should also be perpendicular to z because z is just a translated version of y. So, in this case, the proposition seems to hold true. So, condition ① is correct.Condition ②: x, y are lines, z is a plane.Alright, now x and y are lines, and z is a plane. If x is perpendicular to y, and y is parallel to z, does that mean x is perpendicular to z? Hmm, let's think about this. If y is parallel to plane z, that means y lies within a plane that's parallel to z or something like that. But x is perpendicular to y. Does that necessarily mean x is perpendicular to the entire plane z? I'm not so sure. Because x could be perpendicular to just the line y, but not necessarily to every line in plane z. So, x might not be perpendicular to plane z. In fact, x could be at some angle to plane z, not necessarily 90 degrees. So, condition ② doesn't seem to guarantee the proposition. So, condition ② is incorrect.Condition ③: x is a line, y, z are planes.Okay, now x is a line, and y and z are planes. If x is perpendicular to y, and y is parallel to z, does that mean x is perpendicular to z? Let me recall, if two planes are parallel, then any line perpendicular to one is also perpendicular to the other. That makes sense because parallel planes have the same orientation. So, if x is perpendicular to plane y, and y is parallel to z, then x must also be perpendicular to z. So, condition ③ is correct.Condition ④: z, x, y are all planes.Alright, all three are planes now. If x is perpendicular to y, and y is parallel to z, does that mean x is perpendicular to z? Let me think. If y is parallel to z, then they have the same orientation. If x is perpendicular to y, then it should also be perpendicular to z because z is just a translated version of y in terms of orientation. So, yes, x would be perpendicular to z as well. Therefore, condition ④ is correct.So, summarizing my thoughts:- Condition ①: Correct- Condition ②: Incorrect- Condition ③: Correct- Condition ④: CorrectTherefore, the proposition holds true under conditions ①, ③, and ④.

💡Okay, so I'm trying to solve this problem where there's a rectangle in the x-y plane, and the probability that x + y < 4 is 0.4. I need to find the y-coordinates of two points that have an x-coordinate of 0. Hmm, let's break this down step by step.First, I know that in the x-y plane, a rectangle is defined by its four corners, or vertices. Each vertex has an x and y coordinate. Since the problem mentions points with an x-coordinate of 0, those would be points along the y-axis. So, these two points are somewhere on the y-axis, and I need to find their y-coordinates.Now, the probability that x + y < 4 is 0.4. Probability in this context likely refers to the area where x + y < 4 divided by the total area of the rectangle. So, if I can figure out the total area of the rectangle and the area where x + y < 4, I might be able to find the y-coordinates.Let me visualize this. The line x + y = 4 is a straight line that crosses the x-axis at (4, 0) and the y-axis at (0, 4). Points below this line satisfy x + y < 4. So, if my rectangle is positioned such that part of it is below this line and part is above, the area below the line would be 0.4 times the total area of the rectangle.But wait, I don't know the dimensions of the rectangle. The problem doesn't specify where the rectangle is located or how big it is. That makes it a bit tricky. Maybe I can assume the rectangle is positioned in a standard way? Like, perhaps it's aligned with the axes, meaning its sides are parallel to the x and y axes.If that's the case, then the rectangle would have two points on the y-axis (where x=0) and two points on some other vertical line (where x=a, for some a). Similarly, it would have two points on the x-axis (where y=0) and two points on some other horizontal line (where y=b, for some b). So, the rectangle would have vertices at (0, 0), (a, 0), (a, b), and (0, b).But the problem mentions points with x=0, so those would be (0, 0) and (0, b). Wait, but (0, 0) is a specific point, and (0, b) is another. So, maybe the two points with x=0 are (0, y1) and (0, y2), where y1 and y2 are the y-coordinates I need to find.But then, how does the probability come into play? If the probability that x + y < 4 is 0.4, that means the area of the region within the rectangle where x + y < 4 is 0.4 times the total area of the rectangle.So, let's denote the total area of the rectangle as A = a * b, where a is the length along the x-axis and b is the length along the y-axis.The area where x + y < 4 within the rectangle would depend on where the line x + y = 4 cuts through the rectangle. If the rectangle is entirely below the line, then the entire area would satisfy x + y < 4, making the probability 1. If part of the rectangle is above the line, then only a portion would satisfy x + y < 4.Given that the probability is 0.4, which is less than 1, it means that only 40% of the rectangle's area is below the line x + y = 4.But without knowing the specific dimensions of the rectangle, I can't directly calculate the area below the line. Maybe I need to make some assumptions or set up equations based on the given probability.Let's assume that the rectangle extends from x=0 to x=a and from y=0 to y=b. Then, the line x + y = 4 will intersect the rectangle somewhere. The intersection points would determine the shape of the region where x + y < 4 within the rectangle.If the line intersects the top side of the rectangle (y=b) at some x-coordinate, say x1, and the right side of the rectangle (x=a) at some y-coordinate, say y1, then the region below the line would be a polygon bounded by (0,0), (x1, 0), (a, y1), and (0, b). But this might not always be the case depending on the values of a and b.Alternatively, if the line x + y = 4 passes through the rectangle such that it intersects both the top and right sides, the area below the line would be a trapezoid or a triangle, depending on where the intersections are.This is getting a bit complicated. Maybe I should consider specific cases or try to express the area below the line in terms of a and b.Let's denote the intersection points:1. Intersection with the top side y = b: x + b = 4 ⇒ x = 4 - b2. Intersection with the right side x = a: a + y = 4 ⇒ y = 4 - aNow, depending on the values of a and b, these intersection points may or may not lie within the rectangle.If 4 - b ≤ a, then the intersection with the top side is within the rectangle's x-range (0 to a). Similarly, if 4 - a ≤ b, then the intersection with the right side is within the rectangle's y-range (0 to b).Assuming both intersections are within the rectangle, the area below the line x + y < 4 would be a quadrilateral with vertices at (0,0), (4 - b, 0), (a, 4 - a), and (0, b). But I'm not sure if that's accurate.Wait, actually, if the line intersects both the top and right sides, the area below the line would be a polygon bounded by (0,0), (4 - b, 0), (a, 4 - a), and (0, b). But I need to calculate the area of this polygon.Alternatively, maybe it's easier to calculate the area above the line and subtract it from the total area to find the area below the line.But I'm not sure. Maybe integrating would be a better approach, but since we're dealing with a rectangle and a straight line, perhaps there's a geometric way to find the area.Let me try to sketch this mentally. If the rectangle has vertices at (0,0), (a,0), (a,b), and (0,b), and the line x + y = 4 cuts through it, the area below the line would depend on where the line intersects the rectangle.If the line intersects the top side y = b at x = 4 - b and the right side x = a at y = 4 - a, then the area below the line would consist of two parts: a triangle from (0,0) to (4 - b, 0) to (0, b), and a triangle from (4 - b, 0) to (a, 4 - a) to (0, b). But I'm not sure if that's correct.Alternatively, maybe the area below the line is a trapezoid with vertices at (0,0), (4 - b, 0), (a, 4 - a), and (0, b). The area of a trapezoid is given by the average of the two parallel sides multiplied by the height. But I need to identify the parallel sides.Wait, perhaps it's easier to use the formula for the area of a polygon given its vertices. If I can define the vertices of the region where x + y < 4 within the rectangle, I can use the shoelace formula to calculate the area.But this seems too involved for now. Maybe I should consider specific values or make assumptions about a and b to simplify the problem.Alternatively, perhaps the rectangle is such that the line x + y = 4 passes through two of its vertices. If that's the case, then the area below the line would be exactly half of the rectangle, making the probability 0.5. But since the probability is 0.4, which is less than 0.5, the line must pass closer to one side.Wait, maybe the rectangle is positioned such that the line x + y = 4 intersects two of its sides, not necessarily the top and right sides. It could intersect the top and left sides, or the bottom and right sides, depending on the rectangle's position.But the problem mentions points with x=0, so the rectangle must at least include the y-axis. Therefore, the rectangle extends from x=0 to some x=a, and from y=0 to some y=b.Given that, the line x + y = 4 will intersect the top side y=b at x=4 - b and the right side x=a at y=4 - a.Now, for these intersection points to lie within the rectangle, we must have 0 ≤ 4 - b ≤ a and 0 ≤ 4 - a ≤ b.So, 4 - b ≥ 0 ⇒ b ≤ 4And 4 - a ≥ 0 ⇒ a ≤ 4Also, 4 - b ≤ a ⇒ a ≥ 4 - bAnd 4 - a ≤ b ⇒ b ≥ 4 - aSo, combining these, we have:a ≥ 4 - bb ≥ 4 - aAnd both a and b are positive.This defines a region in the a-b plane where a and b satisfy these inequalities.Now, the area below the line x + y = 4 within the rectangle can be calculated as the area of the rectangle minus the area above the line.The area above the line would be a triangle with vertices at (4 - b, b), (a, 4 - a), and (a, b). The area of this triangle can be calculated using the formula for the area of a triangle given three points.But this is getting too complicated. Maybe there's a simpler way.Alternatively, perhaps I can express the area below the line as a function of a and b and set it equal to 0.4 times the total area, then solve for a and b.Let me try that.Total area of the rectangle, A = a * bArea below the line x + y < 4, A_below = ?If the line intersects the top side at (4 - b, b) and the right side at (a, 4 - a), then the area below the line is the area of the rectangle minus the area of the triangle above the line.The area of the triangle above the line can be calculated as follows:The triangle has vertices at (4 - b, b), (a, 4 - a), and (a, b).The base of the triangle is along the top side from (4 - b, b) to (a, b), which has length a - (4 - b) = a + b - 4.The height of the triangle is along the right side from (a, 4 - a) to (a, b), which has length b - (4 - a) = a + b - 4.Wait, that's interesting. Both the base and the height are equal to a + b - 4.But the area of a triangle is (base * height) / 2, so:Area_triangle = ( (a + b - 4) * (a + b - 4) ) / 2 = ( (a + b - 4)^2 ) / 2Therefore, the area below the line is:A_below = A - Area_triangle = a * b - ( (a + b - 4)^2 ) / 2Given that the probability is 0.4, we have:A_below / A = 0.4 ⇒ (a * b - ( (a + b - 4)^2 ) / 2 ) / (a * b) = 0.4Simplifying:1 - ( (a + b - 4)^2 ) / (2 * a * b ) = 0.4⇒ ( (a + b - 4)^2 ) / (2 * a * b ) = 0.6⇒ (a + b - 4)^2 = 1.2 * a * bNow, this is an equation relating a and b. But we have two variables, so we need another equation to solve for both a and b.However, the problem only asks for the y-coordinates of two points with x=0, which are (0, y1) and (0, y2). In the rectangle I've defined, these points would be (0,0) and (0, b). But (0,0) is a specific point, and the other point is (0, b). So, perhaps the two y-coordinates are 0 and b.But the problem says "two points that have x-coordinate equal to 0," which could mean any two points along the y-axis, not necessarily the vertices of the rectangle. So, maybe the rectangle extends beyond the y-axis, but that complicates things.Alternatively, perhaps the rectangle is such that both points with x=0 are on the y-axis, and the rectangle extends from x=0 to x=a and from y=0 to y=b.In that case, the two points with x=0 would be (0,0) and (0, b). But the problem doesn't specify whether these are vertices or just any points on the y-axis within the rectangle.If they are vertices, then y1=0 and y2=b. But the problem doesn't specify, so I'm not sure.Alternatively, maybe the rectangle is such that the two points with x=0 are not necessarily at the corners but somewhere along the y-axis within the rectangle.But without more information, it's hard to determine.Wait, maybe the rectangle is defined such that the two points with x=0 are the only points on the y-axis, meaning the rectangle is between x=0 and x=a, and between y=y1 and y=y2.In that case, the rectangle would have vertices at (0, y1), (a, y1), (a, y2), and (0, y2).Then, the line x + y = 4 would intersect the rectangle at some points, and the area below the line would be 0.4 times the total area.So, total area A = a * (y2 - y1)The area below the line x + y < 4 would depend on where the line intersects the rectangle.The line x + y = 4 intersects the left side x=0 at y=4, and the right side x=a at y=4 - a.So, if y=4 is within the rectangle's y-range [y1, y2], then the intersection point is (0,4). Similarly, if y=4 - a is within [y1, y2], then the intersection point is (a, 4 - a).Assuming both intersections are within the rectangle, the area below the line would be a trapezoid with vertices at (0, y1), (0,4), (a, 4 - a), and (a, y1).The area of this trapezoid can be calculated as the average of the two parallel sides multiplied by the height.The two parallel sides are along x=0 from y1 to 4, and along x=a from y1 to 4 - a.The lengths of these sides are (4 - y1) and (4 - a - y1).The height is a (the distance along the x-axis).So, area_below = ( (4 - y1) + (4 - a - y1) ) / 2 * a = (8 - 2 y1 - a) / 2 * a = (8 - 2 y1 - a) * a / 2Given that area_below / A = 0.4, and A = a * (y2 - y1), we have:( (8 - 2 y1 - a) * a / 2 ) / (a * (y2 - y1)) ) = 0.4Simplifying:( (8 - 2 y1 - a) / 2 ) / (y2 - y1) = 0.4⇒ (8 - 2 y1 - a) / (2 (y2 - y1)) = 0.4⇒ 8 - 2 y1 - a = 0.8 (y2 - y1)But this introduces another variable, y2, which complicates things further.I think I'm overcomplicating this. Maybe I should consider that the two points with x=0 are the only points on the y-axis, and the rectangle extends from x=0 to x=a and from y=y1 to y=y2.In that case, the line x + y = 4 would intersect the left side at (0,4) and the right side at (a, 4 - a).If (0,4) is within the rectangle, then y1 ≤ 4 ≤ y2.Similarly, if (a, 4 - a) is within the rectangle, then y1 ≤ 4 - a ≤ y2.Assuming both intersections are within the rectangle, the area below the line would be a trapezoid with vertices at (0, y1), (0,4), (a, 4 - a), and (a, y1).The area of this trapezoid is:Area_below = ( (4 - y1) + (4 - a - y1) ) / 2 * a = (8 - 2 y1 - a) / 2 * aThe total area of the rectangle is A = a * (y2 - y1)Given that area_below / A = 0.4, we have:( (8 - 2 y1 - a) / 2 * a ) / (a * (y2 - y1)) ) = 0.4Simplifying:(8 - 2 y1 - a) / (2 (y2 - y1)) = 0.4⇒ 8 - 2 y1 - a = 0.8 (y2 - y1)But we still have two variables, y1 and y2, and only one equation. We need another equation to solve for both variables.Perhaps the rectangle is such that the line x + y = 4 passes through the midpoint of the rectangle? Or maybe the rectangle is symmetric in some way.Alternatively, maybe the rectangle is a square, but the problem doesn't specify that.Without additional information, I think it's impossible to determine the exact y-coordinates of the two points with x=0.Wait, maybe the problem is assuming that the rectangle is such that the two points with x=0 are the only points on the y-axis, and the rectangle is positioned such that the line x + y = 4 divides it into two regions with areas in the ratio 0.4 to 0.6.In that case, perhaps the y-coordinates can be determined based on the intersection points.If the rectangle extends from x=0 to x=a and from y=0 to y=b, and the line x + y = 4 intersects the top side at (4 - b, b) and the right side at (a, 4 - a), then the area below the line is a trapezoid with vertices at (0,0), (4 - b, 0), (a, 4 - a), and (0, b).The area of this trapezoid can be calculated as:Area_below = ( (4 - b) + a ) / 2 * bWait, no, that's not correct. The area of a trapezoid is the average of the two parallel sides multiplied by the height. In this case, the two parallel sides are along x=0 and x=a.At x=0, the side is from y=0 to y=b, so length b.At x=a, the side is from y=0 to y=4 - a, so length 4 - a.The height is a (the distance along the x-axis).So, Area_below = (b + (4 - a)) / 2 * aGiven that area_below / A = 0.4, and A = a * b, we have:( (b + 4 - a) / 2 * a ) / (a * b) ) = 0.4Simplifying:( (b + 4 - a) / 2 ) / b = 0.4⇒ (b + 4 - a) / (2 b) = 0.4⇒ b + 4 - a = 0.8 b⇒ 4 - a = -0.2 b⇒ a = 4 + 0.2 bSo, we have a relation between a and b: a = 4 + 0.2 bBut we need another equation to solve for both a and b.Perhaps the rectangle is such that the line x + y = 4 passes through one of its vertices. For example, if the rectangle's top-right corner is on the line, then a + b = 4.But if a + b = 4, then from a = 4 + 0.2 b, we have:4 + 0.2 b = 4 ⇒ 0.2 b = 0 ⇒ b = 0But b=0 would mean the rectangle collapses to a line, which doesn't make sense. So, that can't be.Alternatively, maybe the line passes through the midpoint of the rectangle. The midpoint would be at (a/2, b/2). Plugging into the line equation:a/2 + b/2 = 4 ⇒ a + b = 8But from earlier, a = 4 + 0.2 bSo, substituting:4 + 0.2 b + b = 8 ⇒ 4 + 1.2 b = 8 ⇒ 1.2 b = 4 ⇒ b = 4 / 1.2 ≈ 3.333Then, a = 4 + 0.2 * 3.333 ≈ 4 + 0.666 ≈ 4.666So, a ≈ 4.666 and b ≈ 3.333But this is just an assumption, and the problem doesn't specify that the line passes through the midpoint.Alternatively, maybe the rectangle is such that the area below the line is a triangle. If the line intersects the rectangle at (0,4) and (4,0), then the area below the line would be a triangle with vertices at (0,0), (0,4), and (4,0).But in this case, the rectangle would have to extend at least to x=4 and y=4, which might not be the case.Given all these uncertainties, I think the problem might be missing some information or requires making specific assumptions.Alternatively, maybe the rectangle is such that the two points with x=0 are the only points on the y-axis, and the rectangle extends from x=0 to x=a and from y=0 to y=b, with the line x + y = 4 cutting through it such that the area below the line is 0.4 times the total area.In that case, using the earlier relation a = 4 + 0.2 b, and knowing that the rectangle must have positive dimensions, we can express the y-coordinates of the two points with x=0 as y=0 and y=b, where b is related to a by a = 4 + 0.2 b.But without another equation, we can't find the exact values of a and b.Wait, maybe the rectangle is such that the line x + y = 4 is tangent to it, meaning it touches the rectangle at exactly one point. In that case, the area below the line would be either entirely inside or outside the rectangle, but that doesn't fit with the probability being 0.4.Alternatively, perhaps the rectangle is positioned such that the line x + y = 4 divides it into two regions with areas in the ratio 0.4 to 0.6, and the two points with x=0 are the intersection points of the line with the y-axis and another point.But I'm not sure.Given all these considerations, I think the problem might be expecting a different approach or might be missing some information.Alternatively, maybe the rectangle is such that the two points with x=0 are the only points on the y-axis, and the rectangle extends from x=0 to x=a and from y=y1 to y=y2, with the line x + y = 4 intersecting the top side at (4 - y2, y2) and the right side at (a, 4 - a).Then, the area below the line would be a trapezoid with vertices at (0, y1), (4 - y2, y2), (a, 4 - a), and (0, y1).But this is getting too convoluted.Perhaps the simplest assumption is that the rectangle is such that the line x + y = 4 intersects the top side at (4 - b, b) and the right side at (a, 4 - a), and the area below the line is 0.4 times the total area.From earlier, we have:a = 4 + 0.2 bSo, if we choose a value for b, we can find a, and vice versa.But without another condition, we can't determine unique values for a and b.Therefore, I think the problem might be expecting us to recognize that the y-coordinates of the two points with x=0 are 0 and 4, but that doesn't make sense because the probability is 0.4, not 0.5.Alternatively, maybe the y-coordinates are such that the area below the line is 0.4 times the total area, leading to specific values for y1 and y2.But without more information, I can't determine the exact y-coordinates.Perhaps the problem is expecting us to recognize that the two points with x=0 are (0,0) and (0,4), but that would make the probability 0.5, not 0.4.Alternatively, maybe the y-coordinates are 0 and some value y such that the area below the line is 0.4 times the total area.But without knowing the rectangle's dimensions, I can't find y.I think I'm stuck here. Maybe I need to look for another approach or see if there's a standard solution to this type of problem.Alternatively, perhaps the rectangle is such that the two points with x=0 are the only points on the y-axis, and the rectangle extends from x=0 to x=a and from y=0 to y=b, with the line x + y = 4 cutting through it such that the area below the line is 0.4 times the total area.In that case, using the earlier relation a = 4 + 0.2 b, and assuming that the rectangle's height b is such that 4 - a ≤ b, we can express b in terms of a.But again, without another equation, we can't solve for both a and b.Alternatively, maybe the rectangle is a square, so a = b.Then, from a = 4 + 0.2 a ⇒ a - 0.2 a = 4 ⇒ 0.8 a = 4 ⇒ a = 5So, a = 5 and b = 5Then, the area below the line would be:Area_below = ( (5 + 4 - 5) / 2 ) * 5 = (4 / 2) * 5 = 2 * 5 = 10Total area A = 5 * 5 = 25Probability = 10 / 25 = 0.4, which matches the given probability.So, if the rectangle is a square with side length 5, then the two points with x=0 are (0,0) and (0,5).But the problem doesn't specify that the rectangle is a square, so this is just an assumption.Alternatively, if the rectangle is not a square, but a = 4 + 0.2 b, and we choose b=5, then a=4 + 0.2*5=4 +1=5, which again gives a square.So, perhaps the only solution is a square with side length 5, making the y-coordinates 0 and 5.But I'm not sure if that's the intended solution.Alternatively, maybe the rectangle is such that the line x + y = 4 passes through the midpoint of the rectangle, leading to a=4 and b=4, but that would make the probability 0.5, not 0.4.Hmm.Given all this, I think the most reasonable assumption is that the rectangle is a square with side length 5, making the y-coordinates of the two points with x=0 equal to 0 and 5.But I'm not entirely confident about this solution.

💡Alright, let's tackle these problems one by one. I'm going to take my time to understand each part and make sure I get it right. **Problem (1): Observing the Sequence**So, we have a sequence where the first term is 3, the second term is 3 squared, the third is 3 cubed, and so on. It says that starting from the second term, the ratio of each term to the previous one is a constant. Hmm, okay, so let me write down the first few terms to visualize it better.- ( a_1 = 3 )- ( a_2 = 3^2 = 9 )- ( a_3 = 3^3 = 27 )- ( a_4 = 3^4 = 81 )- And so on...Now, to find the ratio between consecutive terms, I can divide the second term by the first term: ( frac{a_2}{a_1} = frac{9}{3} = 3 ). Let me check the next ratio: ( frac{a_3}{a_2} = frac{27}{9} = 3 ). Okay, so it seems like the ratio is consistently 3. That makes sense because each term is 3 raised to the power of its term number. So, the ratio is 3.Next, it asks for ( a_6 ). Well, since each term is 3 raised to the power of n, ( a_6 ) should be ( 3^6 ). Let me calculate that: ( 3^6 = 729 ). So, ( a_6 = 729 ).And in general, for any term ( a_n ), it's just ( 3^n ). So, ( a_n = 3^n ).**Problem (2): Finding the Sum of a Geometric Series**Now, this part is about finding the sum ( S_{10} = 1 + 2 + 2^2 + 2^3 + ldots + 2^{10} ). It mentions using a method where we multiply the series by 2 and then subtract the original series from this multiplied series.Let me write down the series:( S_{10} = 1 + 2 + 2^2 + 2^3 + ldots + 2^{10} ) ...(1)Multiplying both sides by 2:( 2S_{10} = 2 + 2^2 + 2^3 + ldots + 2^{11} ) ...(2)Now, subtracting equation (1) from equation (2):( 2S_{10} - S_{10} = (2 + 2^2 + 2^3 + ldots + 2^{11}) - (1 + 2 + 2^2 + 2^3 + ldots + 2^{10}) )On the right side, most terms cancel out:- The ( 2 ) cancels with the ( 2 )- The ( 2^2 ) cancels with the ( 2^2 )- And so on, up until ( 2^{10} ) cancels with ( 2^{10} )What's left is:( S_{10} = 2^{11} - 1 )Calculating ( 2^{11} ) gives 2048, so:( S_{10} = 2048 - 1 = 2047 )So, the sum is 2047.**Problem (3): Applying the Same Method to a Different Series**Now, we have a similar series but with base 3 and 20 terms:( S_{20} = 3 + 3^2 + 3^3 + ldots + 3^{20} )Using the same method as in problem (2), I'll let ( S_{20} = 3 + 3^2 + 3^3 + ldots + 3^{20} ).First, multiply both sides by 3:( 3S_{20} = 3^2 + 3^3 + 3^4 + ldots + 3^{21} )Now, subtract the original series from this multiplied series:( 3S_{20} - S_{20} = (3^2 + 3^3 + ldots + 3^{21}) - (3 + 3^2 + 3^3 + ldots + 3^{20}) )Again, most terms cancel out:- ( 3^2 ) cancels with ( 3^2 )- ( 3^3 ) cancels with ( 3^3 )- And so on, up until ( 3^{20} ) cancels with ( 3^{20} )What remains is:( 2S_{20} = 3^{21} - 3 )So, solving for ( S_{20} ):( S_{20} = frac{3^{21} - 3}{2} )Calculating ( 3^{21} ) is a bit tedious, but we can leave it in exponential form for the answer.**Summary of My Thoughts:**1. For the first problem, recognizing the sequence as a geometric progression with a common ratio of 3 was straightforward. Calculating specific terms and the general term was just applying the formula.2. For the second problem, using the method of multiplying the series by the common ratio and subtracting to find the sum was a standard approach for geometric series. It's a neat trick that simplifies finding the sum without having to add each term individually.3. Applying the same method to the third problem with base 3 was similar, but I had to be careful with the exponents and ensure that the subtraction step correctly cancels out the intermediate terms. It's important to keep track of the first and last terms after subtraction to avoid mistakes.Overall, these problems reinforce the understanding of geometric sequences and series, and the method to find their sums efficiently.

💡First, I need to determine the center of the given circle equation ( x^2 + y^2 + 2x - 4y = 0 ). To do this, I'll rewrite the equation in the standard form of a circle by completing the squares for both ( x ) and ( y ) terms.For the ( x )-terms:[x^2 + 2x = (x + 1)^2 - 1]For the ( y )-terms:[y^2 - 4y = (y - 2)^2 - 4]Substituting these back into the original equation:[(x + 1)^2 - 1 + (y - 2)^2 - 4 = 0][(x + 1)^2 + (y - 2)^2 - 5 = 0][(x + 1)^2 + (y - 2)^2 = 5]From this, the center of the circle is ( (-1, 2) ).Next, I'll substitute the coordinates of the center ( (-1, 2) ) into the line equation ( 3x + y + a = 0 ) to find the value of ( a ):[3(-1) + 2 + a = 0][-3 + 2 + a = 0][-1 + a = 0][a = 1]Therefore, the value of ( a ) is 1.

💡Okay, so I need to figure out what the smallest even abundant number is. Hmm, first of all, what exactly is an abundant number? I think it's a number where the sum of its proper divisors is greater than the number itself. Proper divisors are all the divisors of a number excluding the number itself, right? So, for example, the proper divisors of 6 are 1, 2, and 3, and their sum is 6, which makes 6 a perfect number because the sum equals the number itself.Alright, so an abundant number would be one where this sum is greater than the number. Let me test this with some numbers. Starting from the smallest even numbers since we're looking for the smallest even abundant number.Let's start with 2. Its only proper divisor is 1, and 1 is less than 2, so 2 is definitely not abundant. Next, 4. Its proper divisors are 1 and 2, and their sum is 3, which is still less than 4. So, 4 isn't abundant either.Moving on to 6. As I mentioned earlier, the sum of its proper divisors is 6, making it a perfect number, not abundant. So, 6 is out. Next is 8. The proper divisors of 8 are 1, 2, and 4, and their sum is 7, which is less than 8. So, 8 isn't abundant.Now, 10. Its proper divisors are 1, 2, and 5, and their sum is 8, which is still less than 10. So, 10 isn't abundant either. Hmm, this is taking longer than I thought. Let's try 12.The proper divisors of 12 are 1, 2, 3, 4, and 6. Adding those up: 1 + 2 is 3, plus 3 is 6, plus 4 is 10, and plus 6 is 16. So, the sum of the proper divisors is 16, which is greater than 12. That means 12 is an abundant number.Wait, is 12 even? Yes, it is. So, 12 is an even abundant number. But is it the smallest one? Let me double-check the numbers before 12 to make sure I didn't miss any.We already checked 2, 4, 6, 8, and 10, and none of them were abundant. So, 12 seems to be the first even number that is abundant. Therefore, I think 12 is the smallest even abundant number.Just to be thorough, let me check the next even number after 12, which is 14. Its proper divisors are 1, 2, and 7, and their sum is 10, which is less than 14. So, 14 isn't abundant. The next one is 16. Its proper divisors are 1, 2, 4, and 8, and their sum is 15, which is less than 16. So, 16 isn't abundant either.Continuing, 18. Proper divisors are 1, 2, 3, 6, 9, and their sum is 1 + 2 + 3 + 6 + 9 = 21, which is greater than 18. So, 18 is also abundant, but since 12 is smaller, 12 is still our answer.I think I've checked enough numbers to confirm that 12 is indeed the smallest even abundant number. It's good to verify because sometimes there might be a smaller number that I overlooked, but in this case, all the even numbers below 12 don't meet the criteria of being abundant.