answer：Alright, so I've got this geometry problem here. Let me try to visualize it first. There are two rectangles, EFGH and IJKL. EFGH has sides EF = 8 and EH = 4, so it's a rectangle that's longer in one direction. IJKL has sides IL = 12 and IK = 4, so it's also a rectangle, but longer in the same direction as EFGH since both have a side of length 4. The problem says that GH and IJ are perpendicular. Hmm, okay. So GH is one of the sides of EFGH, and IJ is a side of IJKL. Since they are perpendicular, that probably means that the rectangles are positioned such that one is rotated 90 degrees relative to the other, or maybe they intersect at a right angle.Also, the shaded area within IJKL is equal to three-quarters of the area of EFGH. I need to find the length of EI. EI is probably the distance between points E and I, which are likely corners of the rectangles.Let me start by calculating the areas of both rectangles. For EFGH, the area is EF multiplied by EH, which is 8 times 4. That gives me 32 square units. Now, the shaded area within IJKL is three-quarters of that, so I need to find three-quarters of 32. Let me compute that: (3/4) * 32 = 24. So the shaded area is 24 square units.Since IJKL is a rectangle with sides IL = 12 and IK = 4, its total area is 12 * 4 = 48 square units. The shaded area is 24, which is exactly half of the total area of IJKL. That might be useful information.Now, since GH and IJ are perpendicular, I think that means that the sides GH and IJ form a right angle where they meet. So, if I imagine the two rectangles overlapping, one is positioned such that its side GH is perpendicular to the side IJ of the other rectangle.Given that, I need to figure out how the shaded area relates to the position of EFGH within IJKL. The shaded area is part of IJKL, so it must be the overlapping region between EFGH and IJKL. Since the shaded area is 24, which is half of IJKL's area, that suggests that EFGH covers exactly half of IJKL.Wait, but EFGH has an area of 32, which is larger than half of IJKL's area (which is 24). Hmm, that seems contradictory. Maybe I'm misunderstanding something.Let me re-read the problem. It says the shaded area within IJKL is equal to three-quarters of the area of EFGH. So, the shaded area is 24, which is three-quarters of EFGH's area. That makes sense because 24 is three-quarters of 32. So, the shaded area is the overlapping region between EFGH and IJKL, and it's 24.Since EFGH is a rectangle of 8x4, and IJKL is 12x4, and they are positioned such that GH is perpendicular to IJ, I think that means that one rectangle is placed vertically and the other horizontally relative to each other.Let me try to sketch this mentally. Let's say EFGH is placed such that EF is horizontal and EH is vertical. Then, IJKL is placed such that IL is horizontal and IK is vertical. Since GH and IJ are perpendicular, GH is a vertical side of EFGH, and IJ is a horizontal side of IJKL, so they are indeed perpendicular.Now, the shaded area is the overlapping region. Since the shaded area is 24, which is three-quarters of EFGH's area, that means that three-quarters of EFGH is overlapping with IJKL.Given that, I can think of EFGH being partially inside IJKL, with three-quarters of its area overlapping. So, the overlapping region is a rectangle itself, with area 24.Since both rectangles have a height of 4, the overlapping region must also have a height of 4. Therefore, the width of the overlapping region can be calculated by dividing the area by the height: 24 / 4 = 6. So, the width of the overlapping region is 6 units.Now, since IJKL has a total width of 12, and the overlapping region is 6 units wide, that suggests that EFGH is centered within IJKL or shifted in some way. But since the rectangles are perpendicular, it's more about how they intersect.Wait, maybe I should consider the coordinates. Let me assign coordinates to the points to make it clearer.Let's place point I at the origin (0,0). Then, since IJKL has IL = 12 and IK = 4, point J would be at (12,0), point K at (12,4), and point L at (0,4).Now, EFGH is another rectangle. Let's assume point E is somewhere along the x-axis, say at (x,0). Since EF = 8 and EH = 4, point F would be at (x+8,0), point G at (x+8,4), and point H at (x,4).But wait, the problem says that GH and IJ are perpendicular. GH is the side from G to H, which is vertical, and IJ is the side from I to J, which is horizontal. So, they are naturally perpendicular, which makes sense.Now, the shaded area is the overlapping region between EFGH and IJKL. Since EFGH is placed somewhere along the x-axis, overlapping with IJKL which spans from (0,0) to (12,4).The overlapping region's width would be the distance between the left side of EFGH and the right side of IJKL, but since EFGH is 8 units long, and IJKL is 12 units long, the overlapping width depends on where EFGH is placed.Given that the overlapping area is 24, which is 6 units in width (since height is 4), that means EFGH must be placed such that 6 units of its width overlap with IJKL.So, if EFGH is placed starting at x = a, then the overlapping width is from a to a+8, but since IJKL is from 0 to 12, the overlapping region would be from max(a,0) to min(a+8,12). The length of this overlap is 6 units.So, min(a+8,12) - max(a,0) = 6.Let me solve for a.Case 1: a <= 0. Then, max(a,0) = 0, and min(a+8,12) = a+8. So, a+8 - 0 = 6 => a = -2.Case 2: a > 0. Then, max(a,0) = a, and min(a+8,12) = a+8 (since a+8 <=12 when a <=4). So, a+8 - a = 8, which is not 6. So, this case doesn't work.Case 3: a+8 >12, so min(a+8,12)=12. Then, 12 - a =6 => a=6.So, possible a values are -2 and 6.But since EFGH is a rectangle, and IJKL is from 0 to12, if a=-2, then EFGH would extend from -2 to 6 on the x-axis, overlapping from 0 to6, which is 6 units. Similarly, if a=6, EFGH would extend from6 to14, overlapping from6 to12, which is 6 units.So, EFGH can be placed either starting at x=-2 or x=6.But the problem asks for the length of EI. Point E is at (a,0), and point I is at (0,0). So, the distance EI is |a -0| = |a|.If a=-2, then EI is 2 units. If a=6, then EI is6 units.But wait, the problem says that the shaded area is within IJKL, which is three-quarters of EFGH's area. So, if EFGH is placed at a=-2, then the overlapping area is from0 to6, which is6 units wide, and since the height is4, area is24, which is correct.Similarly, if placed at a=6, overlapping area is6 to12, which is also6 units wide, area24.But the problem might be implying that EFGH is entirely within IJKL? Wait, no, because EFGH has a width of8, and IJKL has a width of12, so if EFGH is placed at a=-2, it extends beyond IJKL on the left, but the overlapping area is still24.But the problem says the shaded area within IJKL is three-quarters of EFGH's area. So, whether EFGH is placed on the left or the right, the overlapping area is24.But the question is asking for the length of EI. If E is at a=-2, then EI is2 units. If E is at a=6, EI is6 units.But which one is it? The problem doesn't specify the direction, but usually, in such problems, unless specified otherwise, we consider the positive direction. So, maybe a=6, making EI=6.Alternatively, maybe both are possible, but since the problem is asking for EI, and in the diagram, probably E is to the right of I, so EI=6.Wait, but let me think again. If EFGH is placed at a=-2, then E is at (-2,0), and I is at (0,0), so EI is2 units. If EFGH is placed at a=6, E is at (6,0), so EI is6 units.But the problem says "the shaded area within IJKL is equal to three-quarters of the area of EFGH." So, regardless of where EFGH is placed, as long as the overlapping area is24, which is three-quarters of32, it's correct.But the problem is asking for the length of EI. So, depending on the placement, it could be2 or6. But since the problem is likely designed to have a unique answer, I think it's6 units.Wait, but let me check the initial assumption. I assumed that the overlapping region is a rectangle of6x4. But since GH and IJ are perpendicular, maybe the overlapping region is a square or something else.Wait, no, since both rectangles have height4, the overlapping region must also have height4, so the width is6.Therefore, the distance EI is either2 or6, but likely6.Wait, but if EFGH is placed at a=6, then E is at (6,0), and I is at (0,0), so EI is6 units. If placed at a=-2, E is at (-2,0), so EI is2 units.But the problem doesn't specify the direction, so maybe both are possible. But in the context of the problem, since IJKL is from0 to12, and EFGH is another rectangle, it's more logical that EFGH is placed to the right of IJKL, making EI=6.Alternatively, maybe the problem is considering the minimal distance, but I think the answer is6 units.Wait, but let me think again. If EFGH is placed at a=6, then the overlapping area is from6 to12, which is6 units, and the area is24. Similarly, if placed at a=-2, overlapping area is from0 to6, which is also6 units, area24.But the problem says "the shaded area within IJKL is equal to three-quarters of the area of EFGH." So, regardless of placement, as long as the overlapping area is24, it's correct.But the question is about EI, the distance from E to I. If E is at6, EI is6. If E is at-2, EI is2. But since the problem is likely designed to have a positive distance, I think EI is6 units.Wait, but maybe I'm overcomplicating. Let me see.The area of EFGH is32, shaded area is24, which is three-quarters. The shaded area is within IJKL, so it's the overlapping part.Since both rectangles have height4, the overlapping width is6. Therefore, EFGH must be shifted such that6 units of its width overlap with IJKL.If EFGH is placed such that its left side is at x=6, then the overlapping region is from6 to12, which is6 units. Therefore, E is at x=6, so EI is6 units.Alternatively, if EFGH is placed such that its right side is at x=6, then E is at x=-2, and EI is2 units.But since the problem doesn't specify the direction, both could be possible. However, in the context of the problem, it's more likely that EFGH is placed to the right of IJKL, making EI=6.Therefore, I think the length of EI is6 units.

answer：Okay, so I have this polar equation: ρ cosθ = 2 sin2θ. Hmm, I need to figure out what kind of curve this represents. The options are a ray and a circle, a line and a circle, two lines, or just a circle. Let me think about how to approach this.First, I remember that in polar coordinates, ρ is the radius and θ is the angle. So, ρ cosθ is actually the x-coordinate in Cartesian coordinates because x = ρ cosθ. That might be useful later. Looking at the equation, ρ cosθ = 2 sin2θ. I also recall that sin2θ is a double angle identity, which can be written as 2 sinθ cosθ. Let me substitute that in:ρ cosθ = 2 * (2 sinθ cosθ)ρ cosθ = 4 sinθ cosθHmm, okay, so both sides have a cosθ term. Maybe I can factor that out or divide both sides by cosθ? But wait, if I divide both sides by cosθ, I have to be careful because cosθ could be zero, which would make the division undefined. So, maybe I should consider two cases: one where cosθ is not zero and another where cosθ is zero.Case 1: cosθ ≠ 0. Then I can divide both sides by cosθ:ρ = 4 sinθOkay, so that's one equation. Now, I need to convert this into Cartesian coordinates to see what kind of curve it is. I know that ρ = 4 sinθ can be rewritten using the identity ρ = 2a sinθ, which represents a circle with radius a and center at (0, a). So, in this case, a = 2, so it's a circle with radius 2 centered at (0, 2). Let me verify that:Starting with ρ = 4 sinθ. Multiply both sides by ρ:ρ² = 4ρ sinθBut ρ² is x² + y², and ρ sinθ is y. So:x² + y² = 4yBring 4y to the left:x² + y² - 4y = 0To write this in standard form, complete the square for y:x² + (y² - 4y + 4) = 4So,x² + (y - 2)² = 4Yes, that's a circle with center at (0, 2) and radius 2. Got it.Case 2: cosθ = 0. When does cosθ equal zero? That happens at θ = π/2 and θ = 3π/2. But in polar coordinates, θ = π/2 is the positive y-axis, and θ = 3π/2 is the negative y-axis. However, in the original equation, if cosθ = 0, then the left side becomes ρ * 0 = 0, and the right side is 2 sin2θ. Let's see what sin2θ is when cosθ = 0.If θ = π/2, then 2θ = π, so sinπ = 0. Similarly, if θ = 3π/2, then 2θ = 3π, and sin3π = 0. So, both sides equal zero. That means for θ = π/2 and θ = 3π/2, any ρ satisfies the equation? Wait, no, because if cosθ = 0, then the equation becomes 0 = 0, which is always true, regardless of ρ. So, that means the entire lines θ = π/2 and θ = 3π/2 are solutions. But in Cartesian coordinates, θ = π/2 is the positive y-axis, which is the line x = 0, y ≥ 0, and θ = 3π/2 is the negative y-axis, which is x = 0, y ≤ 0. So together, they make up the entire y-axis.Wait, but in the original equation, if θ = π/2, then ρ can be any value, right? So, it's not just a single point, but the entire line. So, the equation represents both the circle x² + (y - 2)² = 4 and the line x = 0. So, it's a combination of a circle and a line.Looking back at the options, option B is a line and a circle, which matches my conclusion. Option A is a ray and a circle, but in this case, it's the entire line, not just a ray. So, I think B is the correct answer.Just to make sure I didn't make a mistake, let me recap:1. Started with ρ cosθ = 2 sin2θ.2. Recognized sin2θ = 2 sinθ cosθ, so equation becomes ρ cosθ = 4 sinθ cosθ.3. Factored out cosθ, leading to two cases: cosθ = 0 and ρ = 4 sinθ.4. For cosθ = 0, θ = π/2 or 3π/2, which correspond to the y-axis lines x = 0.5. For ρ = 4 sinθ, converted to Cartesian coordinates and found it's a circle centered at (0, 2) with radius 2.6. Therefore, the equation represents both the circle and the y-axis line.Yes, that seems correct. So, the answer should be B: A line and a circle.**Final Answer**boxed{B}

answer：To determine the value of ( c ), we first solve the first equation for ( x ):[ 3x + 5 = 1 ]Subtract 5 from both sides:[ 3x = -4 ]Divide both sides by 3:[ x = -frac{4}{3} ]Next, we substitute ( x = -frac{4}{3} ) into the second equation:[ cleft(-frac{4}{3}right) + 15 = 3 ]Multiply ( c ) by ( -frac{4}{3} ):[ -frac{4c}{3} + 15 = 3 ]Subtract 15 from both sides:[ -frac{4c}{3} = -12 ]Multiply both sides by ( -3 ):[ 4c = 36 ]Divide both sides by 4:[ c = 9 ]

answer：Okay, so I've got this inequality to prove: [frac{a^{4}}{4a^{4} + b^{4} + c^{4}} + frac{b^{4}}{a^{4} + 4b^{4} + c^{4}} + frac{c^{4}}{a^{4} + b^{4} + 4c^{4}} leqslant frac{1}{2}]given that (a, b, c neq 0). Hmm, interesting. I need to show that the sum of these three fractions is less than or equal to 1/2. Let me try to break this down.First, I notice that each fraction has a similar structure. For example, the first term is (frac{a^{4}}{4a^{4} + b^{4} + c^{4}}). The denominator is 4 times (a^4) plus (b^4) and (c^4). Similarly, the other terms have denominators that are 4 times one variable's fourth power plus the other two variables' fourth powers.I wonder if there's a way to simplify each term or find a common pattern. Maybe I can manipulate each fraction individually to find an upper bound and then sum those bounds.Let me focus on the first term: (frac{a^{4}}{4a^{4} + b^{4} + c^{4}}). I want to find an upper bound for this. Maybe I can compare it to another fraction that's easier to handle. Let's see.If I consider the denominator, (4a^{4} + b^{4} + c^{4}), it's larger than (a^{4} + b^{4} + c^{4}) because of the 4 coefficient on (a^4). So, the denominator is bigger, which would make the whole fraction smaller. But how much smaller?Wait, maybe I can use some inequality here. The AM-GM inequality comes to mind. The Arithmetic Mean-Geometric Mean inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can apply that here.Looking at the denominator (4a^{4} + b^{4} + c^{4}), I can think of it as (a^{4} + a^{4} + a^{4} + a^{4} + b^{4} + c^{4}). So, that's six terms: four (a^4)s, one (b^4), and one (c^4). Using AM-GM on these six terms:[frac{4a^{4} + b^{4} + c^{4}}{6} geq sqrt[6]{a^{4} cdot a^{4} cdot a^{4} cdot a^{4} cdot b^{4} cdot c^{4}}]Simplifying the right side:[sqrt[6]{a^{16} b^{4} c^{4}} = a^{16/6} b^{4/6} c^{4/6} = a^{8/3} b^{2/3} c^{2/3}]Hmm, not sure if that helps directly. Maybe I need a different approach.Another thought: perhaps I can compare each fraction to something simpler. Let's consider the first term again:[frac{a^{4}}{4a^{4} + b^{4} + c^{4}} leq frac{a^{4}}{2a^{4} + 2a^{4}} = frac{a^{4}}{4a^{4}} = frac{1}{4}]Wait, is that valid? Because (4a^{4} + b^{4} + c^{4} geq 4a^{4}), so the denominator is at least (4a^{4}), making the fraction at most (1/4). Similarly, each of the other fractions would also be at most (1/4). So, adding them up, the total would be at most (3/4), which is larger than (1/2). So, that approach doesn't give me a tight enough bound.Maybe I need a better way to bound each fraction. Let me think about the structure of the inequality. Each fraction has a numerator that's one variable's fourth power and a denominator that's four times that variable's fourth power plus the other two variables' fourth powers.What if I consider each fraction as (frac{x^4}{4x^4 + y^4 + z^4}), where (x, y, z) are (a, b, c) in some order. Maybe I can find a relationship between the numerator and the denominator.Another idea: perhaps homogenization. Since all terms are homogeneous of degree 4, maybe I can assume without loss of generality that (a^4 + b^4 + c^4 = 1). That might simplify things.Let me try that. Suppose (a^4 + b^4 + c^4 = 1). Then, the fractions become:[frac{a^{4}}{4a^{4} + b^{4} + c^{4}} = frac{a^4}{4a^4 + (1 - a^4)} = frac{a^4}{3a^4 + 1}]Similarly, the other terms become:[frac{b^{4}}{4b^{4} + a^{4} + c^{4}} = frac{b^4}{3b^4 + 1}][frac{c^{4}}{4c^{4} + a^{4} + b^{4}} = frac{c^4}{3c^4 + 1}]So, the sum becomes:[frac{a^4}{3a^4 + 1} + frac{b^4}{3b^4 + 1} + frac{c^4}{3c^4 + 1}]Hmm, that seems a bit simpler. Now, I need to show that this sum is less than or equal to (1/2).Let me analyze the function (f(t) = frac{t}{3t + 1}). Maybe I can find its properties. Let's compute its derivative to see if it's concave or convex.[f(t) = frac{t}{3t + 1}][f'(t) = frac{(3t + 1)(1) - t(3)}{(3t + 1)^2} = frac{3t + 1 - 3t}{(3t + 1)^2} = frac{1}{(3t + 1)^2}]Since the derivative is positive, the function is increasing. Also, the second derivative:[f''(t) = frac{-6}{(3t + 1)^3}]Which is negative, so the function is concave on its domain.Since (f(t)) is concave, maybe I can apply Jensen's inequality. Jensen's inequality states that for a concave function,[fleft(frac{a + b + c}{3}right) geq frac{f(a) + f(b) + f(c)}{3}]But I need to relate the sum of (f(a^4)), (f(b^4)), and (f(c^4)). Wait, actually, since (f) is concave, the inequality would be:[frac{f(a^4) + f(b^4) + f(c^4)}{3} leq fleft(frac{a^4 + b^4 + c^4}{3}right)]But since we assumed (a^4 + b^4 + c^4 = 1), this becomes:[frac{f(a^4) + f(b^4) + f(c^4)}{3} leq fleft(frac{1}{3}right)]Calculating (f(1/3)):[fleft(frac{1}{3}right) = frac{frac{1}{3}}{3 cdot frac{1}{3} + 1} = frac{frac{1}{3}}{1 + 1} = frac{1}{6}]So,[frac{f(a^4) + f(b^4) + f(c^4)}{3} leq frac{1}{6}][f(a^4) + f(b^4) + f(c^4) leq frac{1}{2}]Which is exactly what we needed to prove! So, by applying Jensen's inequality to the concave function (f(t) = frac{t}{3t + 1}), we've shown that the sum of the fractions is at most (1/2).Wait, let me double-check that. We assumed (a^4 + b^4 + c^4 = 1), which is valid because the inequality is homogeneous. So, scaling (a, b, c) won't affect the inequality. Therefore, the result holds in general.Alternatively, another approach could be to use the Cauchy-Schwarz inequality or other inequalities like Titu's lemma, but I think Jensen's worked out nicely here.Just to make sure, let's test with some specific values. Suppose (a = b = c = 1). Then,[frac{1}{4 + 1 + 1} + frac{1}{1 + 4 + 1} + frac{1}{1 + 1 + 4} = frac{1}{6} + frac{1}{6} + frac{1}{6} = frac{1}{2}]So, equality holds when (a = b = c). That makes sense because in the case of equality, all variables are equal, and the function (f(t)) reaches its maximum sum when all inputs are equal due to concavity.Another test: let (a = 2), (b = 1), (c = 1). Then,First term: (frac{16}{4 cdot 16 + 1 + 1} = frac{16}{64 + 1 + 1} = frac{16}{66} approx 0.2424)Second term: (frac{1}{16 + 4 + 1} = frac{1}{21} approx 0.0476)Third term: (frac{1}{16 + 1 + 4} = frac{1}{21} approx 0.0476)Sum: (0.2424 + 0.0476 + 0.0476 approx 0.3376), which is less than (0.5). So, the inequality holds here as well.What if one variable is much larger? Let (a = 10), (b = 1), (c = 1). Then,First term: (frac{10000}{4 cdot 10000 + 1 + 1} = frac{10000}{40002} approx 0.249975)Second term: (frac{1}{10000 + 4 + 1} = frac{1}{10005} approx 0.00009995)Third term: (frac{1}{10000 + 1 + 4} = frac{1}{10005} approx 0.00009995)Sum: approximately (0.249975 + 0.00009995 + 0.00009995 approx 0.250175), still less than (0.5).So, it seems the inequality holds in these test cases, and the proof using Jensen's inequality seems solid.I think I've got it. By normalizing the variables and applying Jensen's inequality to the concave function, we can show the desired result.

answer：First, let's assume the total worth of the stock is Rs. ( X ).The shopkeeper sells 20% of his stock at a 10% profit:- Worth of 20% stock: ( 0.20X )- Profit from this part: ( 0.10 times 0.20X = 0.02X )The remaining 80% of the stock is sold at a 5% loss:- Worth of 80% stock: ( 0.80X )- Loss from this part: ( 0.05 times 0.80X = 0.04X )The overall loss is given as Rs. 450:[text{Loss from 80% of stock} - text{Profit from 20% of stock} = text{Overall loss}][0.04X - 0.02X = 450][0.02X = 450]Solving for ( X ):[X = frac{450}{0.02} = 22500]Therefore, the total worth of the stock is Rs. ( boxed{22,500} ).

answer：Okay, so I have this problem about an arithmetic sequence. Let me try to understand it step by step. The problem says that in an arithmetic sequence {a_n}, the terms a₄ and a₈ are two roots of the equation x² - 6x + 5 = 0. I need to find the value of a₆, and the options are given as A: ±√5, B: √5, C: ±3, and D: 3.First, I remember that in an arithmetic sequence, each term is obtained by adding a constant difference to the previous term. So, if I denote the first term as a₁ and the common difference as d, then the nth term can be written as a_n = a₁ + (n-1)d.Now, the problem mentions that a₄ and a₈ are roots of the quadratic equation x² - 6x + 5 = 0. I recall that for a quadratic equation of the form x² + bx + c = 0, the sum of the roots is -b and the product is c. So, in this case, the sum of the roots a₄ and a₈ should be equal to 6 (since -(-6) = 6), and the product should be 5.Let me write that down:a₄ + a₈ = 6a₄ * a₈ = 5Since {a_n} is an arithmetic sequence, I can express a₄ and a₈ in terms of a₁ and d:a₄ = a₁ + 3da₈ = a₁ + 7dSo, substituting these into the sum equation:(a₁ + 3d) + (a₁ + 7d) = 6Simplify:2a₁ + 10d = 6Divide both sides by 2:a₁ + 5d = 3Hmm, that's one equation. Now, let's look at the product of the roots:(a₁ + 3d)(a₁ + 7d) = 5I can expand this:(a₁)² + 7a₁d + 3a₁d + 21d² = 5Simplify:(a₁)² + 10a₁d + 21d² = 5But from the earlier equation, I have a₁ + 5d = 3. Maybe I can express a₁ in terms of d:a₁ = 3 - 5dNow, substitute this into the product equation:(3 - 5d)² + 10*(3 - 5d)*d + 21d² = 5Let me compute each term step by step:First, (3 - 5d)²:= 9 - 30d + 25d²Second, 10*(3 - 5d)*d:= 10*(3d - 5d²)= 30d - 50d²Third, 21d² remains as is.Now, add all these together:(9 - 30d + 25d²) + (30d - 50d²) + 21d² = 5Combine like terms:- The constant term: 9- The d terms: -30d + 30d = 0- The d² terms: 25d² - 50d² + 21d² = (25 - 50 + 21)d² = (-24 + 21)d² = -3d²So, the equation simplifies to:9 - 3d² = 5Subtract 5 from both sides:4 - 3d² = 0Which leads to:-3d² = -4Divide both sides by -3:d² = 4/3Take square roots:d = ±2/√3 or d = ±(2√3)/3Hmm, so the common difference d can be either positive or negative 2√3/3.Now, let's find a₁ using a₁ = 3 - 5d.Case 1: d = 2√3/3a₁ = 3 - 5*(2√3/3) = 3 - (10√3)/3Case 2: d = -2√3/3a₁ = 3 - 5*(-2√3/3) = 3 + (10√3)/3So, we have two possible sequences depending on the value of d.But the question asks for a₆. Let's find a₆ in terms of a₁ and d:a₆ = a₁ + 5dUsing the expression for a₁ from earlier, a₁ = 3 - 5d, so:a₆ = (3 - 5d) + 5d = 3Wait, that's interesting. The terms involving d cancel out, leaving a₆ = 3 regardless of the value of d.So, even though d can be positive or negative, a₆ is always 3.Let me double-check this result. If a₄ and a₈ are roots of the equation x² - 6x + 5 = 0, then their sum is 6 and their product is 5. In an arithmetic sequence, the terms equidistant from a₆ are a₄ and a₈, so a₆ is the average of a₄ and a₈. Since a₄ + a₈ = 6, their average is 3, which is a₆. That makes sense.Therefore, the value of a₆ is 3, which corresponds to option D.**Final Answer**boxed{D}