標題:
數學題....唔識
發問:
find the Nth term of the sequence 26 38 50 62 後面個3個係+12.....但第1個點黎
最佳解答:
first term = 26, difference = 12 T(N) = a+(N-1)d, where a is the first term, and d is the difference So, T(N) = 26+(N-1)12 for N = 1,2,3,... [OR T(N) = 14+12N] Check: T(1) = 26+(1-1)12 = 26 T(2) = 26+(2-1)12 = 38 T(3) = 26+(3-1)12 = 50 T(4) = 26+(4-1)12 = 62 For this kind of arithmetic sequence, the first term is always given or have to be found. You can also see a pattern that the difference of any two successive members of the sequence is a constant.
The general term for a sequence with common difference is a + (n-1) * d where a is the first term, n is the term sequence, d is the difference so the Nth term is : 26 + (n-1) * 12 = 26 + 12n - 12 = 12n + 14 testing 1st term : 12 * 1 + 14 = 26 第1個就是咁黎 2nd term : 12 * 2 + 14 = 38 3rd term : 12 * 3 + 14 = 50 4th term : 12 * 4 + 14 = 62
數學題....唔識
發問:
find the Nth term of the sequence 26 38 50 62 後面個3個係+12.....但第1個點黎
最佳解答:
first term = 26, difference = 12 T(N) = a+(N-1)d, where a is the first term, and d is the difference So, T(N) = 26+(N-1)12 for N = 1,2,3,... [OR T(N) = 14+12N] Check: T(1) = 26+(1-1)12 = 26 T(2) = 26+(2-1)12 = 38 T(3) = 26+(3-1)12 = 50 T(4) = 26+(4-1)12 = 62 For this kind of arithmetic sequence, the first term is always given or have to be found. You can also see a pattern that the difference of any two successive members of the sequence is a constant.
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其他解答:The general term for a sequence with common difference is a + (n-1) * d where a is the first term, n is the term sequence, d is the difference so the Nth term is : 26 + (n-1) * 12 = 26 + 12n - 12 = 12n + 14 testing 1st term : 12 * 1 + 14 = 26 第1個就是咁黎 2nd term : 12 * 2 + 14 = 38 3rd term : 12 * 3 + 14 = 50 4th term : 12 * 4 + 14 = 62
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