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A geometric progression maths
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1.The common ratio R of a G.P. is 2 and the sum of the first 6 terms S(6) is 155. Find the first term a. S(6) = 155 a[(2^6)-1]/(2-1) = 155 a(64-1)=155 a = 2.46 (corr to 3 sig fig) 2. C1,C2,C3, is are circles. the radius of C1 is 2cm and the length of radius of the succeeding circle is double of the preceding one. Find the area of C5. am i right? ㄦ=3.141592 Radius of C5 = 2*2^(5-1) = 2*2^4 = 2*16 = 32 Area of C5 = (32)^2 π = 1024π cm^2 2010-04-09 16:02:54 補充： 1.) S(6) = 155 a[(2^7)-1]/(2-1) = 155 a(128-1)=155 a = 1.22(corr to 3 sig fig) 2010-04-09 16:05:28 補充： 補充果到打錯
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A geometric progression maths
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1.The common ratio R of a G.P. is 2 and the sum of the first 6 terms S(6) is 155. Find the first term a.2. C1,C2,C3, is are circles. the radius of C1 is 2cm and the length of radius of the succeeding circle is double of the preceding one. Find the area of C5. am i right?... 顯示更多 1.The common ratio R of a G.P. is 2 and the sum of the first 6 terms S(6) is 155. Find the first term a. 2. C1,C2,C3, is are circles. the radius of C1 is 2cm and the length of radius of the succeeding circle is double of the preceding one. Find the area of C5. am i right? ㄦ=3.141592最佳解答:
1.The common ratio R of a G.P. is 2 and the sum of the first 6 terms S(6) is 155. Find the first term a. S(6) = 155 a[(2^6)-1]/(2-1) = 155 a(64-1)=155 a = 2.46 (corr to 3 sig fig) 2. C1,C2,C3, is are circles. the radius of C1 is 2cm and the length of radius of the succeeding circle is double of the preceding one. Find the area of C5. am i right? ㄦ=3.141592 Radius of C5 = 2*2^(5-1) = 2*2^4 = 2*16 = 32 Area of C5 = (32)^2 π = 1024π cm^2 2010-04-09 16:02:54 補充： 1.) S(6) = 155 a[(2^7)-1]/(2-1) = 155 a(128-1)=155 a = 1.22(corr to 3 sig fig) 2010-04-09 16:05:28 補充： 補充果到打錯
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