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question:Given an arithmetic sequence {a_n}, the sum of the first n terms is denoted as S_n. If S_{10} = 10 and S_{20} = 40, then S_{30} equals ( ) A: 70 B: 90 C: 130 D: 160
answer:Since the sum of the first n terms of the arithmetic sequence {a_n} is S_n, it follows that S_{10}, S_{20} - S_{10}, and S_{30} - S_{20} also form an arithmetic sequence. Therefore, 2(40 - 10) = 10 + S_{30} - 40, Solving this, we get S_{30} = 90, Hence, the correct option is boxed{B}.
question:White tshirts can be purchased in packages of 6. If Mom buys a certain number of packages, she will have 426 white tshirts. How many packages did Mom buy?
answer:To find out how many packages Mom bought, we need to divide the total number of white t-shirts by the number of t-shirts in each package. Total number of white t-shirts = 426 Number of t-shirts in each package = 6 Number of packages = Total number of white t-shirts / Number of t-shirts in each package Number of packages = 426 / 6 Number of packages = 71 Mom bought boxed{71} packages of white t-shirts.
question:Given sets A={3, m^2} and B={1, 3, 2m1}, if A subseteq B, then the value of the real number m is.
answer:If A subseteq B, then we must have m^2 = 2m - 1. The equation m^2 = -1 is discarded. Solving the equation, we get m = 1. Upon verification, it meets the requirement of the distinctness of the elements in the set. Therefore, the answer is boxed{1}.
question:The 51 cards in a deck are numbered 1, 2, cdots, 51. Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked. The two persons with lower numbered cards from a team, and the two persons with higher numbered cards form another team. Let p(a) be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards a and a+7, and Dylan picks the other of these two cards. The minimum value of p(a) for which p(a)gefrac{1}{2} can be written as frac{m}{n}, where m and n are relatively prime positive integers. Find m+n.
answer:Once Alex and Dylan draw their cards, there are binom{49}{2} = 1176 ways for Blair and Corey to draw their cards. Alex and Dylan form the team with higher numbered cards if Blair and Corey both draw below a, which occurs in binom{a-1}{2} ways. Likewise, Alex and Dylan form the team with lower numbered cards if Blair and Corey both draw above a+7, which occurs in binom{42-a}{2} ways. Therefore,[ p(a) = frac{binom{42-a}{2} + binom{a-1}{2}}{1176}. ]Simplifying, we get[ p(a) = frac{(42-a)(41-a)/2 + (a-1)(a-2)/2}{1176}. ]We seek the value of a such that this probability is not lower than frac{1}{2}. Testing several values, it turns out when a is near 22, calculations work effectively due to symmetry around this midpoint. Assume a = 22 + b, then[ p(a) geq frac{1}{2}text{ when }(42 - a)(41 - a) + (a - 1)(a - 2) geq 1176. ]By symmetry argument and checking values, calculating when b = 0 is effective, and testing confirms it meets the criteria. Conclusion: With a = 22 (assuming valid for b = 0), p(22) = frac{820}{1176}, simplifying using gcd finds frac{205}{294}. Therefore, m+n = 205+294 = boxed{499}.