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PP-T1-DS-217

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PP-T1-DS-217

文章lucyyeh » 2007-12-12 15:45

a1 , a2 , a3 , . . . , a15

In the sequence shown, an = an - 1 + k, where 2<=n<=15 and k is a nonzero constant. How many of the terms in the sequence are greater than 10 ?

(1) a1 = 24

(2) a8 = 10

ANS:B
全無頭緒怎麼解的
lucyyeh
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Re: PP-T1-DS-217

文章scottkidd » 2007-12-13 10:37

lucyyeh \$m[1]:a1 , a2 , a3 , . . . , a15

In the sequence shown, an = an - 1 + k, where 2<=n<=15 and k is a nonzero constant. How many of the terms in the sequence are greater than 10 ?

(1) a1 = 24

(2) a8 = 10

ANS:B
全無頭緒怎麼解的

題目說要"greater than 10 ",an又是 = an - 1 + k,k is a nonzero constant。
所以a8=10=a7+k → a7一定<10,a9=a8+k=10+k一定就大於10囉 ;))
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文章a-jan » 2007-12-27 14:21

請問樓主..
(1) 條件不充份的原因,是因為 a1=24 代入 an = an - 1 + k 之下... ==> a1-1= a0 ..因題目只提到 a1.......a15,因此超出題目要求的範圍嗎?

謝..
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文章chris8888 » 2008-01-15 22:33

In the sequence shown, an = an - 1 + k, where 2<=n<=15 and k is a nonzero constant. How many of the terms in the sequence are greater than 10 ?

(1) a1 = 24
a2 = a1 + k => a2 = 24 + k
a3 = a2 + k => a3 = 24 + 2k
a4 = a3 + k => a4 = 24 + 3k
不知到k到底是positive or negative 也不知到 k的大小, 通通會影響>10的個數 insufficient
比方說若k=-15, 那麼全都<10
(2) a8 = 10
a7 = a6 + k => a6 = 10 - 2k
a8 = a7 + k => a7 = 10 - k
a9 = a8 + k => a9 = 10 + k
a10 = a9 + k => a10 = 10 + 2k
a11 = a10 + k => a11 = 10 + 3k
由於已知的數據是10, k不管大小, 都會影響10變成大於10或小於10, 可以推論
由此可見假若k>0, 那麼從a9 ~ a15有七個大於10 ; 反之, 假若k<0, a8 ~ a2有七個大於10
頭像
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Re:

文章william0625 » 2008-08-31 02:08

chris8888 \$m[1]:In the sequence shown, an = an - 1 + k, where 2<=n<=15 and k is a nonzero constant. How many of the terms in the sequence are greater than 10 ?



(1) a1 = 24

a2 = a1 + k => a2 = 24 + k

a3 = a2 + k => a3 = 24 + 2k

a4 = a3 + k => a4 = 24 + 3k

不知到k到底是positive or negative 也不知到 k的大小, 通通會影響>10的個數 insufficient

比方說若k=-15, 那麼全都<10

(2) a8 = 10

a7 = a6 + k => a6 = 10 - 2k

a8 = a7 + k => a7 = 10 - k

a9 = a8 + k => a9 = 10 + k

a10 = a9 + k => a10 = 10 + 2k

a11 = a10 + k => a11 = 10 + 3k

由於已知的數據是10, k不管大小, 都會影響10變成大於10或小於10, 可以推論

由此可見假若k>0, 那麼從a9 ~ a15有七個大於10 ; 反之, 假若k<0, a8 ~ a2有七個大於10


k>0, 那麼從a9 ~ a15有七個大於10 ; 反之, 假若k<0, a8 ~ a2有七個大於10
k>0k<0這樣不就有兩種解
還是只算一種解 可以解釋一下嗎
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Re: PP-T1-DS-217

文章Philosophia » 2008-08-31 17:18

In fact point of (B) is that A8 is in the middle of 15. As a result, when A8 is in the middle of 15 sequence and A8 = 10, it is not important k > 0 or k <0 - there are 7 terms in the sequence are greater than 10.
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Re: PP-T1-DS-217

文章william0625 » 2008-08-31 23:00

Philosophia \$m[1]:In fact point of (B) is that A8 is in the middle of 15. As a result, when A8 is in the middle of 15 sequence and A8 = 10, it is not important k > 0 or k <0 - there are 7 terms in the sequence are greater than 10.


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