中5 Ex2c (5,9,10,14) @ wszspj2的部落格

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中5 Ex2c (5,9,10,14)

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Ex2c (5,9,10,11) http://i293.photobucket.com/albums/mm67/zaza520/0002-97.jpg http://i293.photobucket.com/albums/mm67/zaza520/0003-69.jpg http://i293.photobucket.com/albums/mm67/zaza520/0004-60.jpg

最佳解答:

(2a) First term a = 2, common difference d = 4 No. of terms n = 25 Sum = (n/2)[2a + (n - 1)d] = (25/2) x (4 + 24 x 4) = 1250 (b) First term a = -90, common difference d = 10 No. of terms n = 20 Sum = (n/2)[2a + (n - 1)d] = 10 x (-180 + 19 x 10) = 100 (3a) nth term = a + (n - 1)d 0 + 10(n - 1) = 200 n - 1 = 20 n = 21 (b) First term a = 0, common difference d = 10 No. of terms n = 21 Sum = (n/2)[2a + (n - 1)d] = (21/2) x 20 x 10 = 2100 (6a) 3rd term = a + 2d = 0 ... (1) 6th term = a + 5d = -15 ... (2) Solving, we have d = -5 and a = 10 (b) Sum = (n/2)[2a + (n - 1)d] = 6 x (20 - 5 x 5) = -30 (14a) First term = 1, common difference = -2 20th term = 1 + 19 x (-2) = -37 (b) Sum = (n/2)[2a + (n - 1)d] = 10 x (2 - 19 x 2) = -360 (c) Sum of first 40 terms = 20 x (2 - 39 x 2) = -760 So sum of 21st to 40th term = -760 - (-360) = -400