💡First, I need to find the length of the unknown parallel side of the trapezium. I know the area of the trapezium, one of the parallel sides, and the distance between the two parallel sides.I recall that the formula for the area of a trapezium is:[ text{Area} = frac{1}{2} times (text{Sum of the lengths of the parallel sides}) times (text{Distance between them}) ]Given:- Area = 150 cm²- One parallel side = 20 cm- Distance between the parallel sides = 10 cmLet’s denote the unknown parallel side as ( x ) cm. Plugging the known values into the formula:[ 150 = frac{1}{2} times (20 + x) times 10 ]Simplifying the equation:[ 150 = 5 times (20 + x) ][ 150 = 100 + 5x ][ 50 = 5x ][ x = 10 ]Therefore, the length of the other parallel side is 10 cm.

💡First, I need to determine the total volume of the tank by multiplying its length, width, and height.Next, I'll calculate the effective volume of water each urn can hold by taking 80% of the urn's total volume.Then, I'll find the total volume of water from all the urns by multiplying the number of urns by the effective volume per urn.Finally, I'll determine the percentage of the tank's volume that will be filled by dividing the total water volume by the tank's volume and multiplying by 100%.

💡Okay, so I've been given this problem about a function ( f: [0, infty) to mathbb{R} ) that's continuous and constant on the non-negative integers, ( mathbb{Z}_{geq 0} ). The condition is that for any four points ( 0 leq a < b < c < d ) where ( f(a) = f(c) ) and ( f(b) = f(d) ), it must also hold that ( fleft( frac{a+b}{2} right) = fleft( frac{c+d}{2} right) ). I need to prove that ( f ) is constant everywhere.Alright, let me break this down. First, since ( f ) is constant on ( mathbb{Z}_{geq 0} ), let's denote that constant value as ( c ). So, ( f(n) = c ) for all integers ( n geq 0 ). The function is continuous on the entire interval ( [0, infty) ), which is important because continuity often allows us to extend properties from dense subsets to the entire space.Now, the key condition here is about the midpoints. If I have two pairs of points ( a, c ) and ( b, d ) such that ( f(a) = f(c) ) and ( f(b) = f(d) ), then the function evaluated at the midpoints of these pairs must also be equal. That is, ( fleft( frac{a+b}{2} right) = fleft( frac{c+d}{2} right) ).I wonder if I can use this condition to show that the function must be constant on some dense subset of ( [0, infty) ), and then use continuity to extend it to the entire interval. Maybe starting with the integers and working my way to the dyadic rationals or something like that.Let me try to see what happens if I take specific points. Suppose I take ( a = 0 ) and ( c = 2 ). Since ( f(0) = f(2) = c ), and if I choose ( b = 1 ) and ( d = 3 ), then ( f(1) = f(3) = c ). According to the condition, ( fleft( frac{0+1}{2} right) = fleft( frac{2+3}{2} right) ), so ( f(0.5) = f(2.5) ).Hmm, interesting. So ( f(0.5) = f(2.5) ). But I don't know yet if these are equal to ( c ) or not. Maybe I can create more such points.What if I take ( a = 0.5 ) and ( c = 2.5 ). Then ( f(0.5) = f(2.5) ). If I choose ( b = 1.5 ) and ( d = 3.5 ), then ( f(1.5) = f(3.5) ). Applying the condition again, ( fleft( frac{0.5 + 1.5}{2} right) = fleft( frac{2.5 + 3.5}{2} right) ), which simplifies to ( f(1) = f(3) ). But we already know ( f(1) = f(3) = c ), so this doesn't give us new information.Maybe I need a different approach. Let's think about the function's behavior between integers. Since ( f ) is continuous, if it's constant on a dense set, it must be constant everywhere. So, can I show that ( f ) is constant on all dyadic rationals, which are dense in ( [0, infty) )?Dyadic rationals are numbers of the form ( frac{k}{2^n} ) where ( k ) and ( n ) are integers. If I can show that ( f ) is constant on all dyadic rationals, then by continuity, ( f ) must be constant everywhere.Let's try to use induction on ( n ). For ( n = 0 ), dyadic rationals are integers, and ( f ) is constant there. Suppose ( f ) is constant on all dyadic rationals with denominator ( 2^n ). Now, consider dyadic rationals with denominator ( 2^{n+1} ). These are midpoints between the dyadic rationals of denominator ( 2^n ).If I can show that ( f ) evaluated at these midpoints is equal to the constant value ( c ), then by induction, ( f ) is constant on all dyadic rationals.Wait, how can I apply the given condition to these midpoints? Let's say I have two dyadic rationals ( a ) and ( c ) with denominator ( 2^n ), and their midpoint is ( frac{a + c}{2} ). If ( f(a) = f(c) = c ), then by the given condition, ( fleft( frac{a + c}{2} right) ) must equal ( fleft( frac{c + d}{2} right) ) for some ( d ). Hmm, I'm not sure if this directly helps.Maybe I need to choose specific points ( a, b, c, d ) such that their midpoints are the dyadic rationals I'm interested in. For example, take ( a = 0 ), ( b = frac{1}{2} ), ( c = 1 ), ( d = frac{3}{2} ). Then ( f(a) = f(c) = c ) and ( f(b) = f(d) ) (if ( f ) is constant on these midpoints). Then, the condition tells me that ( fleft( frac{0 + frac{1}{2}}{2} right) = fleft( frac{1 + frac{3}{2}}{2} right) ), which is ( fleft( frac{1}{4} right) = fleft( frac{5}{4} right) ).But I don't know yet if ( fleft( frac{1}{4} right) = c ). Maybe I can create a chain of such equalities. If I keep applying this condition recursively, I might be able to show that all dyadic rationals have the same function value.Alternatively, maybe I can use the fact that ( f ) is continuous and the set of points where ( f ) equals ( c ) is dense. If I can show that ( f(x) = c ) for all ( x ) in a dense subset, then by continuity, ( f ) must be ( c ) everywhere.Let me try to formalize this. Suppose I take any ( x in [0, infty) ). Since the dyadic rationals are dense, there exists a sequence of dyadic rationals ( {x_n} ) such that ( x_n to x ). If I can show that ( f(x_n) = c ) for all ( n ), then by continuity, ( f(x) = lim_{n to infty} f(x_n) = c ).So, the key is to show that ( f ) is equal to ( c ) on all dyadic rationals. Let's try to do this by induction on the denominator ( 2^n ).**Base Case:** For ( n = 0 ), the dyadic rationals are integers, and ( f ) is constant ( c ) there. So, the base case holds.**Inductive Step:** Assume that for all dyadic rationals with denominator ( 2^n ), ( f ) is equal to ( c ). Now, consider a dyadic rational ( frac{k}{2^{n+1}} ). We need to show that ( fleft( frac{k}{2^{n+1}} right) = c ).Note that ( frac{k}{2^{n+1}} ) is the midpoint between ( frac{k - 1}{2^n} ) and ( frac{k + 1}{2^n} ). Since ( f ) is equal to ( c ) at both ( frac{k - 1}{2^n} ) and ( frac{k + 1}{2^n} ) by the inductive hypothesis, we can apply the given condition.Let me set ( a = frac{k - 1}{2^n} ), ( b = frac{k}{2^{n+1}} ), ( c = frac{k + 1}{2^n} ), and ( d = frac{k + 2}{2^{n+1}} ). Wait, no, that might not satisfy ( a < b < c < d ). Let me adjust.Actually, let's set ( a = frac{k - 1}{2^n} ), ( b = frac{k}{2^{n+1}} ), ( c = frac{k + 1}{2^n} ), and ( d = frac{k + 2}{2^{n+1}} ). But I need to ensure that ( a < b < c < d ). Let's check:( a = frac{k - 1}{2^n} ), ( b = frac{k}{2^{n+1}} ), ( c = frac{k + 1}{2^n} ), ( d = frac{k + 2}{2^{n+1}} ).Wait, ( b = frac{k}{2^{n+1}} ) is less than ( c = frac{k + 1}{2^n} ) because ( frac{k}{2^{n+1}} = frac{k}{2 cdot 2^n} ) and ( frac{k + 1}{2^n} ) is much larger. Similarly, ( d = frac{k + 2}{2^{n+1}} ) is less than ( c ) if ( k + 2 < 2(k + 1) ), which is true for ( k geq 0 ).Wait, no, ( d = frac{k + 2}{2^{n+1}} ) is actually less than ( c = frac{k + 1}{2^n} ) because ( frac{k + 2}{2^{n+1}} = frac{k + 2}{2 cdot 2^n} ) and ( frac{k + 1}{2^n} ) is twice that. So, ( a < b < c < d ) holds.Now, ( f(a) = f(c) = c ) by the inductive hypothesis, and ( f(b) = f(d) ) because ( b ) and ( d ) are symmetric around ( c ). Wait, but I don't know if ( f(b) = f(d) ) yet. Hmm, maybe I need a different approach.Alternatively, consider that ( frac{k}{2^{n+1}} ) is the midpoint between ( frac{k - 1}{2^n} ) and ( frac{k + 1}{2^n} ). So, set ( a = frac{k - 1}{2^n} ), ( c = frac{k + 1}{2^n} ), and choose ( b ) and ( d ) such that ( f(b) = f(d) ). Maybe set ( b = frac{k}{2^{n+1}} ) and ( d = frac{k + 2}{2^{n+1}} ). Then, ( f(a) = f(c) = c ), and if I can ensure ( f(b) = f(d) ), then by the given condition, ( fleft( frac{a + b}{2} right) = fleft( frac{c + d}{2} right) ).But ( frac{a + b}{2} = frac{frac{k - 1}{2^n} + frac{k}{2^{n+1}}}{2} = frac{2(k - 1) + k}{2^{n+2}}} = frac{3k - 2}{2^{n+2}}} ). Similarly, ( frac{c + d}{2} = frac{frac{k + 1}{2^n} + frac{k + 2}{2^{n+1}}}{2} = frac{2(k + 1) + k + 2}{2^{n+2}}} = frac{3k + 4}{2^{n+2}}} ).This doesn't seem helpful because I don't know the values of ( f ) at these new points. Maybe I'm complicating things.Let me think differently. Since ( f ) is constant on the integers, and the given condition relates midpoints, perhaps I can use the fact that the function's behavior is constrained by these midpoint equalities, forcing it to be constant everywhere.Another idea: suppose ( f ) is not constant. Then there exist points where ( f(x) neq c ). But since ( f ) is continuous, the set where ( f(x) neq c ) would have to be intervals, but the midpoint condition might prevent such intervals from existing.Wait, maybe I can use the fact that if ( f ) is not constant, then there's some interval where it's different from ( c ). But the midpoint condition would force it to be equal to ( c ) at the midpoints, leading to a contradiction.Alternatively, consider that the function is constant on a dense set, so by continuity, it must be constant everywhere. But I need to show that the function is constant on a dense set, like the dyadic rationals.Let me try to formalize this. Suppose I define ( S ) as the set of all dyadic rationals. I need to show that ( f(x) = c ) for all ( x in S ). Then, since ( S ) is dense and ( f ) is continuous, ( f ) must be ( c ) everywhere.To show ( f(x) = c ) for all ( x in S ), I can use induction on the denominator ( 2^n ).**Base Case:** For ( n = 0 ), ( S ) includes integers, and ( f(x) = c ) there.**Inductive Step:** Assume ( f(x) = c ) for all ( x in S ) with denominator ( 2^n ). Now, consider ( x = frac{k}{2^{n+1}} ). We need to show ( f(x) = c ).Note that ( x ) is the midpoint between ( frac{k - 1}{2^n} ) and ( frac{k + 1}{2^n} ). Let ( a = frac{k - 1}{2^n} ), ( c = frac{k + 1}{2^n} ), ( b = x ), and ( d ) be another point such that ( f(b) = f(d) ). Wait, I need to choose ( d ) such that ( f(d) = f(b) ).Alternatively, since ( f(a) = f(c) = c ), and if I can choose ( b ) and ( d ) such that ( f(b) = f(d) ), then by the given condition, ( fleft( frac{a + b}{2} right) = fleft( frac{c + d}{2} right) ).But I'm stuck on how to choose ( b ) and ( d ) such that ( f(b) = f(d) ). Maybe I need to use the fact that ( f ) is constant on the integers and work from there.Wait, perhaps I can use the fact that ( f ) is constant on the integers to create a chain of equalities. For example, starting from ( f(0) = c ), I can use the midpoint condition to show ( f(0.5) = c ), then use ( f(0.5) = c ) to show ( f(0.25) = c ), and so on.Let me try that. Start with ( a = 0 ), ( c = 2 ), so ( f(a) = f(c) = c ). Choose ( b = 1 ), ( d = 3 ), so ( f(b) = f(d) = c ). Then, by the condition, ( fleft( frac{0 + 1}{2} right) = fleft( frac{2 + 3}{2} right) ), which gives ( f(0.5) = f(2.5) ).But I don't know if ( f(0.5) = c ) yet. However, I can use ( f(0.5) = f(2.5) ) and apply the condition again. Let me set ( a = 0.5 ), ( c = 2.5 ), ( b = 1.5 ), ( d = 3.5 ). Then ( f(a) = f(c) ) and ( f(b) = f(d) ) (since ( f(1.5) = f(3.5) ) if ( f ) is constant on these midpoints). Then, ( fleft( frac{0.5 + 1.5}{2} right) = fleft( frac{2.5 + 3.5}{2} right) ), which simplifies to ( f(1) = f(3) ). But we already know ( f(1) = f(3) = c ), so this doesn't help.Maybe I need to use a different set of points. Let's take ( a = 0 ), ( b = 0.5 ), ( c = 1 ), ( d = 1.5 ). Then ( f(a) = f(c) = c ) and ( f(b) = f(d) ) (if ( f(0.5) = f(1.5) )). Then, by the condition, ( fleft( frac{0 + 0.5}{2} right) = fleft( frac{1 + 1.5}{2} right) ), which is ( f(0.25) = f(1.25) ).If I can show that ( f(0.25) = c ), then by induction, I can extend this to all dyadic rationals. But how?Wait, maybe I can use the fact that ( f ) is continuous. If I can show that ( f ) is constant on a sequence converging to any point, then by continuity, ( f ) must be constant everywhere.Alternatively, consider that the function's value at any dyadic rational can be related back to the integers through repeated midpoint conditions. Since the function is constant on the integers, and the midpoints must match, perhaps this forces the function to be constant on all dyadic rationals.Let me try to formalize this. Suppose I have a dyadic rational ( x = frac{k}{2^n} ). I can express ( x ) as the midpoint of two integers ( a ) and ( c ) such that ( f(a) = f(c) = c ). Then, by the given condition, ( f(x) ) must equal ( f ) evaluated at another midpoint, which can be related back to ( c ).Wait, but ( x ) might not be the midpoint of two integers unless ( k ) is odd. If ( k ) is even, then ( x ) is an integer. If ( k ) is odd, then ( x ) is the midpoint between two integers. So, for ( x = frac{k}{2^n} ), if ( k ) is even, ( x ) is an integer, and ( f(x) = c ). If ( k ) is odd, ( x ) is the midpoint between ( frac{k - 1}{2^n} ) and ( frac{k + 1}{2^n} ), which are both dyadic rationals with denominator ( 2^n ).By the inductive hypothesis, ( fleft( frac{k - 1}{2^n} right) = fleft( frac{k + 1}{2^n} right) = c ). Then, applying the given condition with ( a = frac{k - 1}{2^n} ), ( c = frac{k + 1}{2^n} ), and choosing ( b = x ), ( d ) such that ( f(b) = f(d) ), we get ( f(x) = fleft( frac{c + d}{2} right) ).But I still need to ensure ( f(d) = f(b) ). Maybe I can choose ( d ) symmetrically around ( c ). For example, if ( b = x ), then ( d ) could be ( 2c - x ). Then, ( f(d) = f(b) ) if ( f ) is symmetric around ( c ), but I don't know that yet.Alternatively, since ( f ) is constant on the integers, and the midpoints are forced to have equal values, perhaps this symmetry propagates through all dyadic rationals.Wait, maybe I can use the fact that ( f ) is continuous and the set where ( f(x) = c ) is dense. If I can show that ( f(x) = c ) for all dyadic rationals, then by continuity, ( f ) must be ( c ) everywhere.So, to summarize my thoughts:1. ( f ) is constant on integers.2. The given condition relates midpoints of points where ( f ) is equal.3. By using induction on dyadic rationals, I can show that ( f ) is constant on all dyadic rationals.4. Since dyadic rationals are dense in ( [0, infty) ) and ( f ) is continuous, ( f ) must be constant everywhere.I think this is the right approach. Now, let me try to write this more formally.

💡To determine the value of ((a - b)^3), we first need to find the values of (a) and (b).**Step 1: Calculate (a)**(a) is the number of positive multiples of 8 that are less than 60. - The multiples of 8 less than 60 are: 8, 16, 24, 32, 40, 48, 56.- Counting these, we find there are 7 multiples.Thus, (a = 7).**Step 2: Calculate (b)**(b) is the number of positive integers less than 60 that are multiples of both 4 and 2. - Since every multiple of 4 is also a multiple of 2, we only need to count the multiples of 4.- The multiples of 4 less than 60 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56.- Counting these, we find there are 14 multiples.However, since we are looking for numbers that are multiples of both 4 and 2, and every multiple of 4 is already a multiple of 2, we need to consider only the multiples of 4. Therefore, (b = 14).**Step 3: Compute ((a - b)^3)**- Substitute the values of (a) and (b): (a - b = 7 - 14 = -7).- Cube the result: ((-7)^3 = -343).Thus, the final answer is (boxed{-343}).

💡Okay, so I have this problem here: I need to find the sum of the first six terms in the geometric sequence that starts with 1/4, then 1/16, then 1/64, and so on. Hmm, geometric sequence, right? I remember that in a geometric sequence, each term is found by multiplying the previous term by a common ratio. So, let me see if I can figure out what that ratio is.First, the first term is 1/4, and the second term is 1/16. So, if I divide the second term by the first term, that should give me the common ratio, r. Let me calculate that: (1/16) divided by (1/4). Dividing fractions is the same as multiplying by the reciprocal, so that would be (1/16) * (4/1) which equals 4/16, which simplifies to 1/4. Okay, so the common ratio r is 1/4.Now, I need to find the sum of the first six terms. I remember there's a formula for the sum of the first n terms of a geometric sequence. Let me recall what that formula is. I think it's S_n = a * (1 - r^n) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms. Let me write that down:S₆ = a * (1 - r⁶) / (1 - r)Plugging in the values I have: a is 1/4, r is 1/4, and n is 6. So substituting these into the formula:S₆ = (1/4) * (1 - (1/4)⁶) / (1 - 1/4)Let me compute each part step by step. First, let's calculate (1/4)⁶. That's (1/4) multiplied by itself six times. Let me compute that:(1/4)² = 1/16(1/4)³ = 1/64(1/4)⁴ = 1/256(1/4)⁵ = 1/1024(1/4)⁶ = 1/4096Okay, so (1/4)⁶ is 1/4096. Now, let's compute 1 - (1/4)⁶, which is 1 - 1/4096. To subtract these, I can write 1 as 4096/4096, so:1 - 1/4096 = 4096/4096 - 1/4096 = 4095/4096Alright, so the numerator in the formula becomes (1/4) * (4095/4096). Let me compute that:(1/4) * (4095/4096) = 4095 / (4 * 4096) = 4095 / 16384Wait, hold on, is that right? Let me double-check. 4 * 4096 is 16384, yes. So, (1/4) * (4095/4096) is indeed 4095/16384.Now, the denominator in the formula is (1 - 1/4). Let's compute that:1 - 1/4 = 3/4So, putting it all together, the formula becomes:S₆ = (4095/16384) / (3/4)Dividing by a fraction is the same as multiplying by its reciprocal, so:(4095/16384) * (4/3) = (4095 * 4) / (16384 * 3)Let me compute the numerator and the denominator separately.First, the numerator: 4095 * 4. Let me calculate that:4000 * 4 = 16,00095 * 4 = 380So, 16,000 + 380 = 16,380Wait, that doesn't seem right. Wait, 4095 * 4. Let me do it step by step:4000 * 4 = 16,00090 * 4 = 3605 * 4 = 20So, 16,000 + 360 = 16,360 + 20 = 16,380. Okay, that's correct.Now, the denominator: 16384 * 3. Let me compute that:16,000 * 3 = 48,000384 * 3 = 1,152So, 48,000 + 1,152 = 49,152So, putting it together, we have:16,380 / 49,152Now, let's simplify this fraction. Let's see if 16,380 and 49,152 have any common factors.First, let's check if both are divisible by 2:16,380 ÷ 2 = 8,19049,152 ÷ 2 = 24,576So, 8,190 / 24,576Check if both are divisible by 2 again:8,190 ÷ 2 = 4,09524,576 ÷ 2 = 12,288So, 4,095 / 12,288Check if both are divisible by 3:4,095 ÷ 3 = 1,36512,288 ÷ 3 = 4,096So, now we have 1,365 / 4,096Let me check if 1,365 and 4,096 have any common factors. 1,365 is an odd number, and 4,096 is a power of 2 (2^12), so they don't have any common factors besides 1. So, the simplified fraction is 1,365 / 4,096.Wait a second, but earlier I had 16,380 / 49,152, which simplified to 1,365 / 4,096. Let me just confirm that:16,380 ÷ 12 = 1,36549,152 ÷ 12 = 4,096Yes, that's correct. So, the sum S₆ is 1,365 / 4,096.Wait, but earlier in my initial calculation, I had 4095 / 16384 divided by 3/4, which gave me 16,380 / 49,152, which simplifies to 1,365 / 4,096. So, that seems consistent.But let me cross-verify this result another way to make sure I didn't make a mistake.Alternatively, I can compute each term up to the sixth term and add them up manually.First term: 1/4 = 0.25Second term: 1/16 = 0.0625Third term: 1/64 = 0.015625Fourth term: 1/256 ≈ 0.00390625Fifth term: 1/1024 ≈ 0.0009765625Sixth term: 1/4096 ≈ 0.000244140625Now, let's add these up:0.25 + 0.0625 = 0.31250.3125 + 0.015625 = 0.3281250.328125 + 0.00390625 = 0.332031250.33203125 + 0.0009765625 ≈ 0.33300781250.3330078125 + 0.000244140625 ≈ 0.333251953125Now, let's convert 1,365 / 4,096 to a decimal to see if it matches.1,365 ÷ 4,096. Let me compute that:4,096 goes into 1,365 zero times. So, 0.Add a decimal point and a zero: 13,650 ÷ 4,096.4,096 goes into 13,650 three times (3 * 4,096 = 12,288). Subtract: 13,650 - 12,288 = 1,362.Bring down another zero: 13,620 ÷ 4,096.4,096 goes into 13,620 three times (3 * 4,096 = 12,288). Subtract: 13,620 - 12,288 = 1,332.Bring down another zero: 13,320 ÷ 4,096.4,096 goes into 13,320 three times (3 * 4,096 = 12,288). Subtract: 13,320 - 12,288 = 1,032.Bring down another zero: 10,320 ÷ 4,096.4,096 goes into 10,320 two times (2 * 4,096 = 8,192). Subtract: 10,320 - 8,192 = 2,128.Bring down another zero: 21,280 ÷ 4,096.4,096 goes into 21,280 five times (5 * 4,096 = 20,480). Subtract: 21,280 - 20,480 = 800.Bring down another zero: 8,000 ÷ 4,096.4,096 goes into 8,000 one time (1 * 4,096 = 4,096). Subtract: 8,000 - 4,096 = 3,904.Bring down another zero: 39,040 ÷ 4,096.4,096 goes into 39,040 nine times (9 * 4,096 = 36,864). Subtract: 39,040 - 36,864 = 2,176.Bring down another zero: 21,760 ÷ 4,096.4,096 goes into 21,760 five times (5 * 4,096 = 20,480). Subtract: 21,760 - 20,480 = 1,280.Bring down another zero: 12,800 ÷ 4,096.4,096 goes into 12,800 three times (3 * 4,096 = 12,288). Subtract: 12,800 - 12,288 = 512.Bring down another zero: 5,120 ÷ 4,096.4,096 goes into 5,120 one time (1 * 4,096 = 4,096). Subtract: 5,120 - 4,096 = 1,024.Bring down another zero: 10,240 ÷ 4,096.4,096 goes into 10,240 two times (2 * 4,096 = 8,192). Subtract: 10,240 - 8,192 = 2,048.Bring down another zero: 20,480 ÷ 4,096.4,096 goes into 20,480 five times (5 * 4,096 = 20,480). Subtract: 20,480 - 20,480 = 0.So, putting it all together, 1,365 / 4,096 ≈ 0.333251953125Which matches the decimal sum I calculated earlier: approximately 0.333251953125So, that confirms that the fraction 1,365 / 4,096 is correct.Therefore, the sum of the first six terms is 1,365 / 4,096.Wait, but earlier in my initial calculation, I had 4095 / 16384 divided by 3/4, which gave me 16,380 / 49,152, which simplifies to 1,365 / 4,096. So, that's consistent.But let me just make sure I didn't make any calculation errors in my initial formula application.Starting again:S₆ = a * (1 - r⁶) / (1 - r)a = 1/4, r = 1/4, n = 6So,S₆ = (1/4) * (1 - (1/4)⁶) / (1 - 1/4)Compute (1/4)⁶ = 1/4096So, 1 - 1/4096 = 4095/4096Then, (1/4) * (4095/4096) = 4095 / 16384Denominator: 1 - 1/4 = 3/4So, S₆ = (4095 / 16384) / (3/4) = (4095 / 16384) * (4/3) = (4095 * 4) / (16384 * 3) = 16,380 / 49,152Simplify numerator and denominator by dividing both by 12: 16,380 ÷ 12 = 1,365; 49,152 ÷ 12 = 4,096So, S₆ = 1,365 / 4,096Yes, that's correct.Alternatively, I can also compute the sum by adding each term:Term 1: 1/4 = 0.25Term 2: 1/16 = 0.0625Term 3: 1/64 = 0.015625Term 4: 1/256 ≈ 0.00390625Term 5: 1/1024 ≈ 0.0009765625Term 6: 1/4096 ≈ 0.000244140625Adding them up:0.25 + 0.0625 = 0.31250.3125 + 0.015625 = 0.3281250.328125 + 0.00390625 = 0.332031250.33203125 + 0.0009765625 ≈ 0.33300781250.3330078125 + 0.000244140625 ≈ 0.333251953125Now, let's convert 1,365 / 4,096 to decimal:1,365 ÷ 4,096 ≈ 0.333251953125Which matches the sum we got by adding each term. So, that's a good check.Therefore, I'm confident that the sum of the first six terms is 1,365 / 4,096.But just to make sure, let me try another approach. Maybe using the formula for the sum of a geometric series in another way.I know that the sum Sₙ = a₁ + a₂ + ... + aₙ, where each aₖ = a * r^(k-1)So, for n=6, S₆ = a * (1 - r⁶) / (1 - r)Which is the same formula I used earlier.So, plugging in the values again:a = 1/4, r = 1/4S₆ = (1/4) * (1 - (1/4)⁶) / (1 - 1/4)Compute (1/4)⁶ = 1/4096So, 1 - 1/4096 = 4095/4096Multiply by 1/4: (1/4) * (4095/4096) = 4095 / 16384Divide by (1 - 1/4) = 3/4: (4095 / 16384) / (3/4) = (4095 / 16384) * (4/3) = (4095 * 4) / (16384 * 3) = 16,380 / 49,152Simplify: Divide numerator and denominator by 12: 16,380 ÷ 12 = 1,365; 49,152 ÷ 12 = 4,096So, S₆ = 1,365 / 4,096Same result as before.I think I've verified it sufficiently. So, the sum of the first six terms is 1,365 over 4,096.Just to write it as a fraction in the simplest form, 1,365 and 4,096 have no common factors besides 1, as 1,365 is divisible by 5 (since it ends with a 5), but 4,096 is 2^12, so it's not divisible by 5. Therefore, the fraction is already in its simplest form.So, the final answer is 1,365/4,096.**Final Answer**The sum of the first six terms is boxed{dfrac{1365}{4096}}.

💡First, I need to determine the number of sections in the apple orchard. I know that each section produces 45 sacks of apples daily, and the total daily harvest is 360 sacks.To find the number of sections, I can divide the total number of sacks by the number of sacks produced per section.So, 360 sacks divided by 45 sacks per section equals 8 sections.Therefore, there are 8 sections in the orchard.