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F.5 Arithmetic sequence一問
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請幫我解答,無限感激!請詳盡列式,plz!1.The sum of the first three terms of an Arithmetic sequence is 36.The second term is 3 times od the third term.Find the first three mumbers of the sequence.2. Thefirst term of an Arithmetic sequence is 11 and the common difference is 6. Find the 3 consecutive terms of the sequence such... 顯示更多 請幫我解答,無限感激!請詳盡列式,plz! 1.The sum of the first three terms of an Arithmetic sequence is 36.The second term is 3 times od the third term.Find the first three mumbers of the sequence. 2. Thefirst term of an Arithmetic sequence is 11 and the common difference is 6. Find the 3 consecutive terms of the sequence such that their sum is 195. 3. In an Arithmetic sequence, the sum of 3 consecutive terms is 30 nd their product is 910. Find the possible values of these three terms.
1.The sum of the first three terms of an Arithmetic sequence is 36.The second term is 3 times od the third term.Find the first three mumbers of the sequence. T(1) + T(2) + T(3) = 36 3a + 3d = 36 a+d = 12 --------------(1) a+d = 3(a+2d) a+d = 3a + 6d -2a = 5d 2a = -5d --------------(2) (1)*2 - (2), 2d = 24+5d => d = -8 Sub d = -8 into (1), => a = 20 Hence, the first three numbers are 20, 20-8, 20-16 => 20,12,4 2. Thefirst term of an Arithmetic sequence is 11 and the common difference is 6. Find the 3 consecutive terms of the sequence such that their sum is 195. a = 11, d = 6 Let the one in the middle of the three numbers be a + nd, a + (n-1)d + a + nd + a + (n+1)d = 195 3a + 3nd = 195 a + nd = 65 11 + 6n = 65 6n = 54 n = 9 Hence, the three numbers are 11+(9-1)6, 11+9(6), 11+(9+1)6 => 59,65,71 3. In an Arithmetic sequence, the sum of 3 consecutive terms is 30 nd their product is 910. Find the possible values of these three terms. Let the middle of the three terms be a, a-d + a + a+d = 30 3a = 30 a = 10 (a-d)a(a+d) = 910 (a^2 - d^2)a = 910 100 - d^2 = 91 d^2 = 9 d = +/-3 Hence, this three numbers are 7,10,13
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F.5 Arithmetic sequence一問
發問:
請幫我解答,無限感激!請詳盡列式,plz!1.The sum of the first three terms of an Arithmetic sequence is 36.The second term is 3 times od the third term.Find the first three mumbers of the sequence.2. Thefirst term of an Arithmetic sequence is 11 and the common difference is 6. Find the 3 consecutive terms of the sequence such... 顯示更多 請幫我解答,無限感激!請詳盡列式,plz! 1.The sum of the first three terms of an Arithmetic sequence is 36.The second term is 3 times od the third term.Find the first three mumbers of the sequence. 2. Thefirst term of an Arithmetic sequence is 11 and the common difference is 6. Find the 3 consecutive terms of the sequence such that their sum is 195. 3. In an Arithmetic sequence, the sum of 3 consecutive terms is 30 nd their product is 910. Find the possible values of these three terms.
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最佳解答:1.The sum of the first three terms of an Arithmetic sequence is 36.The second term is 3 times od the third term.Find the first three mumbers of the sequence. T(1) + T(2) + T(3) = 36 3a + 3d = 36 a+d = 12 --------------(1) a+d = 3(a+2d) a+d = 3a + 6d -2a = 5d 2a = -5d --------------(2) (1)*2 - (2), 2d = 24+5d => d = -8 Sub d = -8 into (1), => a = 20 Hence, the first three numbers are 20, 20-8, 20-16 => 20,12,4 2. Thefirst term of an Arithmetic sequence is 11 and the common difference is 6. Find the 3 consecutive terms of the sequence such that their sum is 195. a = 11, d = 6 Let the one in the middle of the three numbers be a + nd, a + (n-1)d + a + nd + a + (n+1)d = 195 3a + 3nd = 195 a + nd = 65 11 + 6n = 65 6n = 54 n = 9 Hence, the three numbers are 11+(9-1)6, 11+9(6), 11+(9+1)6 => 59,65,71 3. In an Arithmetic sequence, the sum of 3 consecutive terms is 30 nd their product is 910. Find the possible values of these three terms. Let the middle of the three terms be a, a-d + a + a+d = 30 3a = 30 a = 10 (a-d)a(a+d) = 910 (a^2 - d^2)a = 910 100 - d^2 = 91 d^2 = 9 d = +/-3 Hence, this three numbers are 7,10,13
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