標題:
f 5 sequence ~ help!!
Show that 32 can be divided into four parts T(1), T(2), T(3) and T(4) which are in arithmetic sequence and such that T(1) x T(4) : T(2) x T(3) is 7 : 15 . thanks for your help~
最佳解答:
Let a be first term = T(1), d be common difference T(2) = a + d, T(3) = a + 2d, T(4) = a + 3d 32 = (a)(a+d)(a+2d)(a+3d) ------------(1) a(a+3d) / (a+d)(a+2d) = 7/15 15a^2 + 45ad = 7a^2 +21ad+ 14d^2 8a^2 + 24ad - 14d^2 = 0 2(2a-d)(2a+7d) = 0 a = 0.5d or a = -3.5d sub a = 0.5 d into (1): 32 = 0.5d(0.5d+d)(0.5d+2d)(0.5d+3d) 32 = 0.5d(1.5d)(2.5d)(3.5d) d^4 = 512/105 d = (512/105)^(1/4) or -(512/105)^(1/4) a = 0.5d= (32/105)^(1/4) or a = - (312/105)^(1/4) or sub a = -3.5d into (1): 32 = (-3.5d)(-3.5d+d)(-3.5d+2d)(-3.5d+3d) d^4 = 512/105 d = (512/105)^(1/4) or -(512/105)^(1/4) same result as the case a = 0.5d
其他解答:
so answer is either a = 0.74300232 and d = 1.48600464 OR a = 5.20101624 and d = -1.48600464
f 5 sequence ~ help!!
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發問:Show that 32 can be divided into four parts T(1), T(2), T(3) and T(4) which are in arithmetic sequence and such that T(1) x T(4) : T(2) x T(3) is 7 : 15 . thanks for your help~
最佳解答:
Let a be first term = T(1), d be common difference T(2) = a + d, T(3) = a + 2d, T(4) = a + 3d 32 = (a)(a+d)(a+2d)(a+3d) ------------(1) a(a+3d) / (a+d)(a+2d) = 7/15 15a^2 + 45ad = 7a^2 +21ad+ 14d^2 8a^2 + 24ad - 14d^2 = 0 2(2a-d)(2a+7d) = 0 a = 0.5d or a = -3.5d sub a = 0.5 d into (1): 32 = 0.5d(0.5d+d)(0.5d+2d)(0.5d+3d) 32 = 0.5d(1.5d)(2.5d)(3.5d) d^4 = 512/105 d = (512/105)^(1/4) or -(512/105)^(1/4) a = 0.5d= (32/105)^(1/4) or a = - (312/105)^(1/4) or sub a = -3.5d into (1): 32 = (-3.5d)(-3.5d+d)(-3.5d+2d)(-3.5d+3d) d^4 = 512/105 d = (512/105)^(1/4) or -(512/105)^(1/4) same result as the case a = 0.5d
其他解答:
so answer is either a = 0.74300232 and d = 1.48600464 OR a = 5.20101624 and d = -1.48600464
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