Urgent! Arithmetic sequence－gbu94mz95e的部落格

標題:

Urgent! Arithmetic sequence

發問:

13. If the sum of the arithmetic sequence-10, ... , 38 is 182,find the number of terms of the sequence.14. How many consecutive positive integers, beginning at 1, should be taken to make the sum equal to 2016?27. The interior angles of a polygon form an arithmetic sequence. The largest angle is 106? and the... 顯示更多 13. If the sum of the arithmetic sequence-10, ... , 38 is 182,find the number of terms of the sequence.14. How many consecutive positive integers, beginning at 1, should be taken to make the sum equal to 2016?27. The interior angles of a polygon form an arithmetic sequence. The largest angle is 106? and the common difference is 5?(a) Find the number of sides of the polygon.(b) Find the smallest interior angle.Please solve the questions step by step. Thank you very much! 更新: To Tsui Q27(a) has soiution, n is 9. 更新 2: To Tsui Yes,the largest angle should be 160?in Q27,I made careless mistake.Thank you for your great help.

最佳解答:

13. Ifthe sumofthe arithmetic sequence -10, ... , 38 is 182, find thenumber of terms of the sequence. Solution : T(1) = -10 T(n) = 38 S(n): n[T(1) + (T(n)]/2 = 182 n[(-10) + 38]/2 = 182 14n = 182 n = 13 Hence, no. of terms = 13 14. Howmany consecutive positive integers, beginning at 1, should be taken to make thesum equal to 2016? Solution : 1 + 2 + 3 + ...... + n = 2016 T(1) = a = 1 d = 1 S(n): n[2a + (n - 1)d]/2 = 2016 n[2*1 + (n - 1)*1]/2 = 2016 n[2 + (n - 1)] = 2016*2 n2 + n - 4032 = 0 (n - 63)(n + 64) = 0 n = 63 or n = -64 (rejected) Hence, no. of consecutive positive integers = 63 27. Theinterior angles of a polygon form an arithmetic sequence. The largest angle is106? and the common differenceis 5? (a)Find the number of sides of the polygon. (b)Find the smallest interior angle. Solution : T(1) = a d = 5 T(n) : a + (n - 1)d = 106 a + (n - 1)*5 = 106 a + 5n - 5 = 106 a = 111 - 5n ...... [1] S(n): n[T(1) + T(n)]/2 = (2n - 4)*90 n(a + 106)/2 = 180n - 360 n(a + 106) = 360n - 720 ...... [2] Put [1] into [2] : n(111 - 5n + 106) = 360n - 720 n(217 - 5n) = 360n - 720 217n - 5n2 = 360n - 720 5n2 + 143n - 720 = 0 n = (-143 ± √34849)/10 n must be a positive integer. Hence, there is no solution. 2011-09-18 19:01:53 補充： In Q3, it should be "160°" instead of "106°" ! (a) T(1) = a d = 5 T(n) : a + (n - 1)*5 = 160 a = 165 - 5n ...... [1] S(n): n(a + 160)/2 = 180n - 360 n(a + 160) = 360n - 720 ...... [2] 2011-09-18 19:02:03 補充： Put [1] into [2] : n(165 - 5n + 160) = 360n - 720 5n2 + 35n - 720 = 0 n2 + 7n - 144 = 0 (n + 16)(n - 9) = 0 n = -16 (rejected) or n = 9 ...... (answer)