answer:Let the common difference of the arithmetic sequences {a_n} and { sqrt {S_n}} be d. Then, a_n = a_1 + (n-1)d, quad sqrt {S_n} = sqrt {a_1} + (n-1)d From the given information, we have sqrt {a_1 + a_2} = sqrt {2a_1 + d} = sqrt {a_1} + d, quad sqrt {3a_1 + 3d} = sqrt {a_1} + 2d Squaring both sides and simplifying, we obtain a_1 + d = d^2 + 2sqrt{a_1}d, quad 2a_1 + 3d = 4d^2 + 4sqrt{a_1}d Solving these equations yields a_1 = 2sqrt{a_1}d - d^2 Substituting this back into the first equation, we get d(2d - 1) = 0 Hence, d = 0 or d = frac{1}{2}. If d = 0, then a_1 = 0, which does not satisfy the condition of positive terms. Thus, we discard this case. Therefore, d = frac{1}{2} and a_1 = frac{1}{4}. Consequently, a_6 = a_1 + (6-1)d = frac{1}{4} + frac{1}{2} cdot 5 = boxed{frac{11}{4}}

answer:1. Given the set (S = {a, b, c, d}) contains the elements (a, b, c), and (d) and satisfies the property that for any (x, y in S), (xy in S). 2. We have the conditions: [ a^2 = 1, quad b^2 = 1, quad c^2 = b ] 3. From (a^2 = 1) and (b^2 = 1), we infer: [ a in {1, -1} quad text{and} quad b in {1, -1} ] 4. Since (S) is closed under multiplication, we analyze the value for (b): - If (b = 1), then (c^2 = 1). This means (c in {1, -1}). - If (c = 1), then (c) and (b) being both 1 does not impose further restrictions on (d). Checking the condition (c = 1): [ d in S, , d cdot c = d cdot 1 = d in S ] - However, if (c = -1), then we need (d cdot c = d cdot (-1) = -d in S), which does not necessarily contradict anything. 5. However, if (b = 1), the closeness under multiplication fails to give convincing set values with the restrictions we have, and complications arise with (c^2 = b). 6. Thus, assume (b = -1): - Then, (c^2 = -1). Therefore, (c) must be: [ c in {mathrm{i}, -mathrm{i} } ] - If (c = mathrm{i}), then we reason with each condition: [ d = -mathrm{i} ] - If (c = -mathrm{i}), then: [ d = mathrm{i} ] Since (d) and (c) just swap places, (S) will still be closed under multiplication. 7. Hence, if we have (b = -1), (c) and (d) would be ({mathrm{i}, -mathrm{i}}). 8. Therefore: [ b + c + d = -1 + mathrm{i} - mathrm{i} = -1 ] # Conclusion: [ boxed{-1} ]

answer:According to the given information, we have sin beta = frac{1}{2}sin alpha (equation 1). Since tan alpha = 3 tan beta, we can rewrite this as frac{sin alpha}{cos alpha} = frac{3 sin beta}{cos beta}, which gives us cos beta = frac{3}{2} cos alpha (equation 2), or sin alpha = 0 (equation 3). If equation 2 holds, we can square both equations 1 and 2, and add them together to get 1 = frac{1}{4}sin^2 alpha + frac{9}{4}cos^2 alpha = frac{1}{4} + 2cos^2 alpha. Solving for cos^2 alpha, we find that cos^2 alpha = frac{3}{8}. Using the double angle formula for cosine, we obtain cos 2alpha = 2cos^2 alpha - 1 = -frac{1}{4}. If equation 3 holds, then cos^2 alpha = 1, and cos 2alpha = 2cos^2 alpha - 1 = 1. Therefore, the possible solutions are boxed{cos 2alpha = -frac{1}{4}} or boxed{cos 2alpha = 1}. To find cos^2 alpha, we use equations 1 and 2 by squaring and adding them. If sin alpha = 0, we have another value for cos^2 alpha. Finally, we apply the double angle formula to find the value of cos 2alpha. This problem primarily tests your understanding of trigonometric identities, the relationships between trigonometric functions, and basic algebraic manipulations.

answer:Let Z be the meeting point of Alex and Blake. Since d = rt, XZ = 10t and YZ = 9t. Applying the Law of Cosines to triangle XYZ: [ YZ^2 = XZ^2 + XY^2 - 2 cdot XZ cdot XY cdot cos(45^circ) ] [ (9t)^2 = (10t)^2 + 150^2 - 2 cdot 10t cdot 150 cdot cos(45^circ) ] [ 81t^2 = 100t^2 + 22500 - 3000t cdot frac{sqrt{2}}{2} ] [ 19t^2 - 1500sqrt{2}t + 22500 = 0 ] Solving this quadratic equation for t: [ t = frac{1500sqrt{2} pm sqrt{(1500sqrt{2})^2 - 4 cdot 19 cdot 22500}}{2 cdot 19} ] [ t = frac{1500sqrt{2} pm sqrt{4500000 - 1710000}}{38} ] [ t = frac{1500sqrt{2} pm sqrt{2790000}}{38} ] Choose the smaller t for the earliest meeting. Calculate t, then determine the distance Alex has cycled: [ XZ = 10t ] Box the final answer: [ boxed{10t text{ meters}} ]