answer:To solve this problem, we need to calculate the relative speed of the two trains when they are moving towards each other and then use the time it takes for them to clear each other to find the length of the second train. When two objects are moving towards each other, their relative speed is the sum of their individual speeds. In this case, the speeds are 42 kmph and 30 kmph. First, we convert the speeds from kmph to m/s by multiplying by (1000 m / 1 km) and dividing by (3600 s / 1 hour): Speed of first train in m/s = 42 kmph * (1000 m / 1 km) / (3600 s / 1 hour) = 42 * 1000 / 3600 = 11.67 m/s (approximately) Speed of second train in m/s = 30 kmph * (1000 m / 1 km) / (3600 s / 1 hour) = 30 * 1000 / 3600 = 8.33 m/s (approximately) Now, we add the speeds to get the relative speed: Relative speed = 11.67 m/s + 8.33 m/s = 20 m/s The time it takes for the trains to clear each other is given as 23.998 seconds. The distance covered by both trains in this time is the sum of their lengths. We can use the formula: Distance = Relative speed * Time Let's denote the length of the second train as L2. The total distance covered by both trains as they clear each other is the length of the first train (200 m) plus the length of the second train (L2). Distance = 200 m + L2 Now we can plug in the values for relative speed and time to find the distance: Distance = 20 m/s * 23.998 s = 479.96 m (approximately) Now we can solve for L2: 479.96 m = 200 m + L2 L2 = 479.96 m - 200 m L2 = 279.96 m Therefore, the length of the second train is approximately boxed{279.96} meters.

answer:1. To prove the statement, we start with the definition of the harmonic mean. If h is the harmonic mean of numbers a and b, then: [ h = frac{2ab}{a + b} ] By taking the reciprocal, we find: [ frac{1}{h} = frac{a + b}{2ab} ] 2. Next, note that (frac{1}{h}) is the arithmetic mean of (frac{1}{a}) and (frac{1}{b}), because: [ frac{1}{h} = frac{a + b}{2ab} = frac{1}{2} left( frac{1}{a} + frac{1}{b} right) ] This reformulates our problem in terms of arithmetic means rather than harmonic means. 3. The problem can now be stated as: "Prove that it is not possible to construct an infinite sequence of natural numbers, where not all elements are equal, and every element (from the second one onward) is the arithmetic mean of its two neighboring elements." 4. Let's consider the sequence of the reciprocals of those natural numbers. Suppose we have a sequence of natural numbers (a_1, a_2, a_3, ldots) such that for ( i geq 2 ): [ a_i = frac{2a_{i-1}a_{i+1}}{a_{i-1} + a_{i+1}} ] 5. Considering the sequence (left{ frac{1}{a_i} right}), we now need to work with: [ frac{1}{a_i} = frac{frac{1}{a_{i-1}} + frac{1}{a_{i+1}}}{2} ] 6. This sequence (left{ frac{1}{a_i} right}) forms an arithmetic sequence. Our goal is to show that an infinite arithmetic sequence of reciprocals of natural numbers, where not all numbers are equal, cannot exist. 7. Reciprocals of natural numbers lie within the interval [0, 1]. Suppose (frac{1}{a}) and (frac{1}{b}) are two consecutive terms of the sequence, where (a) and (b) are natural numbers. Because the sequence forms an arithmetic progression, the difference between consecutive terms is a constant non-zero value: [ d = frac{1}{a_{i+1}} - frac{1}{a_i} ] 8. For a non-constant sequence, this difference (d neq 0). However, in an arithmetic progression, this difference (d) being non-zero implies that the sequence (frac{1}{a_1}, frac{1}{a_2}, frac{1}{a_3}, ldots) eventually must pass beyond the interval [0, 1] for a sufficiently large index, which is impossible for natural number reciprocals. 9. Hence no such infinite sequence of distinct natural numbers exists such that the second onwards elements are the harmonic means of their neighbouring elements. 10. **Conclusion:** The statement is thus proven that no such harmonic mean-based sequence exists in the realm of natural numbers. [ boxed{} ]

answer:1. Define sets ( A_n ) and ( B_n ) where ( A_n ) is the set of integer sequences ( a_1, a_2, ldots, a_n ) that satisfy the given conditions and ( B_n subset A_n ) is the set of sequences where at least one of ( a_2, ldots, a_n ) is zero. We aim to find ( f(n) = |A_n| ) and ( g(n) = |B_n| ). 2. The initial conditions are given as: [ f(1) = 1, quad g(1) = 0, quad f(2) = 2, quad g(2) = 1 ] 3. Consider the set difference ( A_n setminus B_n ). By definition, sequences in ( A_n setminus B_n ) have ( a_2, ldots, a_n ) all greater than 0. Since ( a_1 = 0 ) and ( a_2 leq u_2 + 1 = 1 ), we have ( a_2 = 1 ). Thus, the sequence ( a_3, ldots, a_n ) reports a sequence in ( A_{n-1} ) with each term increased by 1. Conversely, we can derive all sequences in ( A_n setminus B_n ) from sequences in ( A_{n-1} ) by adding 1 to each term and appending an initial 0. Therefore: [ f(n) - g(n) = f(n-1) ] 4. Next, we analyze ( B_n ) for ( n geq 3 ) by partitioning it into three classes: - **Class 1:** sequences where ( a_n = 0 ) or ( 1 ), and at least one element in ( a_2, ldots, a_{n-1} ) is zero. Removing ( a_n ), these seqiences correspond two-to-one to sequences in ( B_{n-1} ). Thus, there are ( 2g(n-1) ) such sequences. - **Class 2:** sequences where ( a_n = 0 ) and ( a_2, ldots, a_{n-1} ) are all greater than 0. Removing ( a_n ), these correspond one-to-one to sequences in ( A_{n-1} setminus B_{n-1} ), which has size ( f(n-1) - g(n-1) = f(n-2) ). - **Class 3:** sequences where ( a_n geq 2 ) and at least one element in ( a_2, ldots, a_{n-1} ) is zero. Assume the last zero element is at position ( k ). Removing ( a_2, ldots, a_{k-1} ) (all zeros), these correspond one-to-one to sequences ending with a term not smaller than 2 in ( A_{n-k+1} setminus B_{n-k+1} ), eventually corresponding to ( A_{n-k} setminus B_{n-k} ) by reindexing. Summing over all ( k ) yields ( (f(n-2) - f(n-3)) + (f(n-3) - f(n-4)) + cdots + (f(2) - f(1)) ). Since ( f(1) = 1 ), this sum provides ( f(n-2) - 1 ). 5. Combining the partitions, we derive: [ g(n) = 2g(n-1) + f(n-2) + (f(n-2) - 1) = 2g(n-1) + 2f(n-2) - 1 ] 6. Eliminating ( g(n) ) from the recursive dependencies, we get: [ f(n) = 3f(n-1) - 1 ] 7. Solving this recurrence relation with initial condition ( f(1) = 1 ): - ( f(2) = 2 ) - Assume ( f(n) = frac{3^{n-1} + 1}{2} ) holds true for some ( n ), proving for ( n+1 ): [ f(n+1) = 3f(n) - 1 = 3left(frac{3^{n-1}+1}{2}right) - 1 = frac{3^n + 3}{2} - 1 = frac{3^n + 1}{2} ] - Therefore, by induction, ( f(n) = frac{3^{n-1} + 1}{2} ). 8. Conclusion: [ boxed{frac{3^{n-1}+1}{2}} ]

answer:Since EG is the angle bisector of angle E, by the Angle Bisector Theorem, we find that frac{DG}{GE} = frac{DF}{EF} = frac{3}{4}. Thus, DG < GE, making GE the longer subsegment of DE. Let GE = x. Then, DG = frac{3x}{4}. We know that DG + GE = DE = 12 inches. Substituting our values, we get: frac{3x}{4} + x = 12 Rightarrow frac{7x}{4} = 12 Rightarrow x = frac{48}{7}. Thus, GE = boxed{frac{48}{7}} inches.