標題:
發問:
Given a geometric sequence whose sum of the first 10 terms us 4 and whose sum from the 11th to the 30th is 48, find the sum from the 31st to the 60th term. Please give detailed solution! 更新: * us --> is
最佳解答:
SUM OF GEOMETRIC SEQUENCE = a(r^n - 1)/(r - 1) So, a(r^10 - 1)/(r - 1) = 4...(1) ar^10(r^20 - 1)/(r - 1) = 48...(2) So, 4/(r^10 - 1) = 48/[r^10(r^20 - 1)] r^10(r^10 + 1) = 12 r^10 = 3 sum from the 31st to the 60th = ar^30(r^30 - 1)/(r - 1) = 2*27*26 = 1404 =
其他解答:
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Geometric sequence發問:
Given a geometric sequence whose sum of the first 10 terms us 4 and whose sum from the 11th to the 30th is 48, find the sum from the 31st to the 60th term. Please give detailed solution! 更新: * us --> is
最佳解答:
SUM OF GEOMETRIC SEQUENCE = a(r^n - 1)/(r - 1) So, a(r^10 - 1)/(r - 1) = 4...(1) ar^10(r^20 - 1)/(r - 1) = 48...(2) So, 4/(r^10 - 1) = 48/[r^10(r^20 - 1)] r^10(r^10 + 1) = 12 r^10 = 3 sum from the 31st to the 60th = ar^30(r^30 - 1)/(r - 1) = 2*27*26 = 1404 =
其他解答:
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