標題:
geometric sequence
發問:
最佳解答:
1. 4 = a(r^10 - 1)/(r - 1) ................(1) 4 + 48 = 52 = a(r^30 - 1)/(r - 1) .................(2) (2)/(1) we get 52/4 = (r^30 - 1)/(r^10 - 1) = 13 Let r^10 = x, we get 13 = (x^3 - 1)/(x - 1) 13(x - 1) = x^3 - 1 x^3 - 13x + 12 = 0 By Factor theorem, x = 1 is a factor, so the equation can be factorized as (x - 1)(x - 3)(x + 4) = 0 so x = r^10 = 1, 3 or - 4. r^10 = 1 and - 4 are rejected, because r cannot be equal to 1 and r must be real. so only solution is r^10 = 3.......(4) Sub. into (1), we get a/(r - 1) = 4/(r^10 - 1) = 4/(3 - 1) = 2.......(5) Using the result of (4) and (5), Sum to 60th term = a(r^60 - 1)/(r - 1) = [a/(r - 1)][(r^10)^6 - 1] = 2(3^6 - 1) = 2(729 - 1) = 1456 So sum of 31st term to 60th term = 1456 - 4 - 48 = 1404.
其他解答:
Let a be the first term and d be the common difference. Sum of the first 10 terms: T(1) + T(2) + T(3) + ...... + T(10) = 4 10[2a + (10 - 1)d]/2 = 4 10a + 45d = 4 ...... (1) Sum from the 11th term to the 30 term: T(11) + T(12) + T(13) + ...... T(30) = 48 [T(1) + T(2) + ...... + T(30)] - [T(1) + T(2) + ...... T(10)] = 48 30[2a + (30 - 1)d]/2 - 10[2a + (10 - 1)d]/2 = 48 30a + 435d - 10a - 45d = 48 20a + 390d = 48 ...... (2) (2) - 2*(1) : 300d = 40 d = 2/15 Put d = 4/30 into (1): 10a + 45(2/15) = 4 10a + 6 = 4 10a = -2 a = -1/5 The sum from 31st to the 60th term = T(31) + T(32) + ...... + T(60) = [T(1) + T(2) + T(3) + ...... T(60)] - [T(1) + T(2) + T(3) + ...... + T(30)] = {60[2*(-1/5) + (60 - 1)*(2/15)]/2} - 48 = {30[(-2/5) + (118/15)]} - 48 = 30*(112/15) - 48 = 224 - 48 = 176
geometric sequence
發問:
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1. Given a geometric sequence whose sum of the first 10 terms is 4 and whose sum from the 11th to the 30th term is 48, find the sum from the 31st to the 60th term. 唔好跳steps~最佳解答:
1. 4 = a(r^10 - 1)/(r - 1) ................(1) 4 + 48 = 52 = a(r^30 - 1)/(r - 1) .................(2) (2)/(1) we get 52/4 = (r^30 - 1)/(r^10 - 1) = 13 Let r^10 = x, we get 13 = (x^3 - 1)/(x - 1) 13(x - 1) = x^3 - 1 x^3 - 13x + 12 = 0 By Factor theorem, x = 1 is a factor, so the equation can be factorized as (x - 1)(x - 3)(x + 4) = 0 so x = r^10 = 1, 3 or - 4. r^10 = 1 and - 4 are rejected, because r cannot be equal to 1 and r must be real. so only solution is r^10 = 3.......(4) Sub. into (1), we get a/(r - 1) = 4/(r^10 - 1) = 4/(3 - 1) = 2.......(5) Using the result of (4) and (5), Sum to 60th term = a(r^60 - 1)/(r - 1) = [a/(r - 1)][(r^10)^6 - 1] = 2(3^6 - 1) = 2(729 - 1) = 1456 So sum of 31st term to 60th term = 1456 - 4 - 48 = 1404.
其他解答:
Let a be the first term and d be the common difference. Sum of the first 10 terms: T(1) + T(2) + T(3) + ...... + T(10) = 4 10[2a + (10 - 1)d]/2 = 4 10a + 45d = 4 ...... (1) Sum from the 11th term to the 30 term: T(11) + T(12) + T(13) + ...... T(30) = 48 [T(1) + T(2) + ...... + T(30)] - [T(1) + T(2) + ...... T(10)] = 48 30[2a + (30 - 1)d]/2 - 10[2a + (10 - 1)d]/2 = 48 30a + 435d - 10a - 45d = 48 20a + 390d = 48 ...... (2) (2) - 2*(1) : 300d = 40 d = 2/15 Put d = 4/30 into (1): 10a + 45(2/15) = 4 10a + 6 = 4 10a = -2 a = -1/5 The sum from 31st to the 60th term = T(31) + T(32) + ...... + T(60) = [T(1) + T(2) + T(3) + ...... T(60)] - [T(1) + T(2) + T(3) + ...... + T(30)] = {60[2*(-1/5) + (60 - 1)*(2/15)]/2} - 48 = {30[(-2/5) + (118/15)]} - 48 = 30*(112/15) - 48 = 224 - 48 = 176
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