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求救! 有關geometric sequence既題目
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1) The first and the last term of a geometric sequence are 2 and 128 respectively. If the sum of all the terms of the sequence is 86, find the common ratio of the sequence. Let r be the common ratio and let n be the number of terms. The first term is a = 2. The last term is ar??1 = 128 ? r??1 = 64 The sum of n terms is a(r? - 1)/(r - 1) = 86 (r? - 1)/(r - 1) = 43 r? - 1 = 43(r - 1) r??1 - 1/r = 43(1 - 1/r) by dividing r on both sides 64 - 1/r = 43 - 43/r 42/r = -21 r = -2 The common ratio of the sequence is r = -2. 2) The common ratio and the sum to infinity of a geometric sequence are 2/3 and 2592 respectively. (a) Find the first term of the sequence. Let a be the first term. It is given that the common ratio is r = 2/3 and the sum to infinity is 2592. a/(1 - r) = 2592 a/(1 - 2/3) = 2592 a/(1/3) = 2592 3a = 2592 a = 864 (b) If the nth term of the sequence is 256,find (i) the value of n, The nth term of the sequence is ar??1 = 256 864(2/3)??1 = 256 (2/3)??1 = 8/27 n-1 = 3 n = 4 (II) the sum of the first n terms. The sum of the first n terms is a(r? - 1)/(r - 1) = (864)[(2/3)? - 1]/(2/3 - 1) = (864)[1 - (2/3)?]/(1 - 2/3) = (864)[1 - (2/3)?]/(1/3) = 2592[1 - (2/3)?] = 2080 3) The sum of the first 5 terms and the sum of the first 10 terms of a geometric sequence are 31 and 1023 respectively. Find the first term and the common ratio of the sequence. Let the first term be a and the common ratio be r. a(r? - 1)/(r - 1) = 31 ...[1] a(r1? - 1)/(r - 1) = 1023 a(r? + 1)(r? - 1)/(r - 1) = 1023 ...[2] [2] ÷ [1] gives r? + 1 = 33 r? = 32 r = 2 Put into [1], a(2? - 1)/(2 - 1) = 31 a(2? - 1) = 31 a(32 - 1) = 31 a(31) = 31 a = 1 Therefore, the first term is a = 1 and the common ratio is r = 2. 2014-01-26 03:35:17 補充： 阿月～ 謝謝你的意見～ ☆ヾ(???)ノ 2014-01-26 03:43:44 補充： 某程度上，我和你也複雜了。 其實用 r^n = 64r 就好了 ^___^
其他解答:
分享我Q1的做法： Let r be the common ratio and n be the number of terms. 2r^(n-1)=128 -(1) 2(1-r^n) / 1-r =86 2-2r^n=86-86r 2r^n=86r-84 -(2) (2)÷(1)： r=(86r-84) / 128 128r=86r-84 84=-42r r=-2 2014-01-26 02:03:50 補充： 第3題原來係 (r^10-1)=(r^5+1)(r^5-1) 諗唔到添... 2014-01-26 09:58:27 補充： 係啵~ Good~
求救! 有關geometric sequence既題目
發問:
此文章來自奇摩知識+如有不便請留言告知
求教這幾題ORZ1) The first and the last term of a geometric sequence are 2 and 128 respectively.If the sum of all the terms of the sequence is 86,find the common ratio of the sequence.2)The common ratio and the sum to infinity of a geometric sequence are 2/3 ang 2592 respectively.(a)Find the first term of the... 顯示更多 求教這幾題ORZ 1) The first and the last term of a geometric sequence are 2 and 128 respectively.If the sum of all the terms of the sequence is 86,find the common ratio of the sequence. 2)The common ratio and the sum to infinity of a geometric sequence are 2/3 ang 2592 respectively. (a)Find the first term of the sequence. (b)If the nth term of the sequence is 256,find (i)the value of n, (II)the sum of the first n terms. 3)The sum of the first 5 terms and the sum of the first 10 terms of a geometric sequence are 31 and 1023 respectively.Find the first term and the common ratio of the sequence. 完全唔明點做......求救ORZ THX!!!!最佳解答:
1) The first and the last term of a geometric sequence are 2 and 128 respectively. If the sum of all the terms of the sequence is 86, find the common ratio of the sequence. Let r be the common ratio and let n be the number of terms. The first term is a = 2. The last term is ar??1 = 128 ? r??1 = 64 The sum of n terms is a(r? - 1)/(r - 1) = 86 (r? - 1)/(r - 1) = 43 r? - 1 = 43(r - 1) r??1 - 1/r = 43(1 - 1/r) by dividing r on both sides 64 - 1/r = 43 - 43/r 42/r = -21 r = -2 The common ratio of the sequence is r = -2. 2) The common ratio and the sum to infinity of a geometric sequence are 2/3 and 2592 respectively. (a) Find the first term of the sequence. Let a be the first term. It is given that the common ratio is r = 2/3 and the sum to infinity is 2592. a/(1 - r) = 2592 a/(1 - 2/3) = 2592 a/(1/3) = 2592 3a = 2592 a = 864 (b) If the nth term of the sequence is 256,find (i) the value of n, The nth term of the sequence is ar??1 = 256 864(2/3)??1 = 256 (2/3)??1 = 8/27 n-1 = 3 n = 4 (II) the sum of the first n terms. The sum of the first n terms is a(r? - 1)/(r - 1) = (864)[(2/3)? - 1]/(2/3 - 1) = (864)[1 - (2/3)?]/(1 - 2/3) = (864)[1 - (2/3)?]/(1/3) = 2592[1 - (2/3)?] = 2080 3) The sum of the first 5 terms and the sum of the first 10 terms of a geometric sequence are 31 and 1023 respectively. Find the first term and the common ratio of the sequence. Let the first term be a and the common ratio be r. a(r? - 1)/(r - 1) = 31 ...[1] a(r1? - 1)/(r - 1) = 1023 a(r? + 1)(r? - 1)/(r - 1) = 1023 ...[2] [2] ÷ [1] gives r? + 1 = 33 r? = 32 r = 2 Put into [1], a(2? - 1)/(2 - 1) = 31 a(2? - 1) = 31 a(32 - 1) = 31 a(31) = 31 a = 1 Therefore, the first term is a = 1 and the common ratio is r = 2. 2014-01-26 03:35:17 補充： 阿月～ 謝謝你的意見～ ☆ヾ(???)ノ 2014-01-26 03:43:44 補充： 某程度上，我和你也複雜了。 其實用 r^n = 64r 就好了 ^___^
其他解答:
分享我Q1的做法： Let r be the common ratio and n be the number of terms. 2r^(n-1)=128 -(1) 2(1-r^n) / 1-r =86 2-2r^n=86-86r 2r^n=86r-84 -(2) (2)÷(1)： r=(86r-84) / 128 128r=86r-84 84=-42r r=-2 2014-01-26 02:03:50 補充： 第3題原來係 (r^10-1)=(r^5+1)(r^5-1) 諗唔到添... 2014-01-26 09:58:27 補充： 係啵~ Good~
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