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標題:
F5 maths A.S.
發問:
If p, q, r and s are 4 consecutive terms of an arithmetic sequence and p+q=8, r+s=24, then the common difference of the arithmetic sequence is A. 2 B. 4 C. 8 D. 12
最佳解答:
Let a be the first term and the common difference is d. p=a,q=a+d,r=a+2d,s=a+3d (It doesn't matter that which is the first term actually) p+q=8 a+(a+d)=8 2a+d=8................(1) r+s=24 (a+2d)+(a+3d)=24 2a+5d=24.............(2) (2)-(1): 4d=16, d=4 The common difference is 4. The answer should be B. 2010-12-23 12:11:52 補充： You can let q=p+d, r=p+2d, s= p+3d, d is the common difference. And the same result is obtained.
其他解答:
F5 maths A.S.
發問:
If p, q, r and s are 4 consecutive terms of an arithmetic sequence and p+q=8, r+s=24, then the common difference of the arithmetic sequence is A. 2 B. 4 C. 8 D. 12
最佳解答:
Let a be the first term and the common difference is d. p=a,q=a+d,r=a+2d,s=a+3d (It doesn't matter that which is the first term actually) p+q=8 a+(a+d)=8 2a+d=8................(1) r+s=24 (a+2d)+(a+3d)=24 2a+5d=24.............(2) (2)-(1): 4d=16, d=4 The common difference is 4. The answer should be B. 2010-12-23 12:11:52 補充： You can let q=p+d, r=p+2d, s= p+3d, d is the common difference. And the same result is obtained.
其他解答:
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p=2 or 6,q=6 or 2,r=10 or 14,s=14 or 10 偶數相加得8的可能結果:2+6,4+4,∵p和q是連續偶數 ∴p和q皆不是4 ∴p=2 or 6,q=6 or 2 偶數相加得24的可能結果:2+22(捨去,因為r or s 不會等於2) 4+20(捨去,因為r or s 不會等於4) 6+18(捨去,因為r or s 不會等於6) 12+12(捨去,因為r 和 s 不相等) 8+16(捨去,因為連續規律不等於p和q) 10+14 ∴r=10 or 14,s= 14 or 10 2010-12-23 12:59:14 補充： ∴答案為B
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