math AS

-75, -63, -51, ..., -3
S(n) is the min
When it consists of all negative number in the sequence.
Required S(n) = -75 + (-63) + (-51) + ... +(-3) = (-75-3) x 7 /2 = -273
The corresponding value of n is 7.

2014-12-28 01:09:40 補充：
To know the last negative number
We should know the general term T(n) first
T(n)=-75+(n-1)(12)= 12n-87
T(n)<=0
12n-87<=0
n<=7.25
N is a positive integer, so n=7
The last negative number is located at n=7.
The last negative number is T(7) = 12(7)-87 = -3

泉路 同學，你是正確的，但如果是計算出來的會比試數的好。

請加入補充，我會刪除我的答案。

2014-12-28 00:50:43 補充：
a = first term = -75
d = common difference = -63 - (-75) = 12

Sum of n terms is
S(n) = (n/2) [2a + (n - 1)d]
　　= (n/2) [-150 + 12(n - 1)]
　　= (n/2) (12n - 162)
　　= (n) (6n - 81)
　　= 6n² - 81n

2014-12-28 00:50:53 補充：
Method 1
S(n) would be minimum when it adds up all the negative terms.

Consider T(n) = a + (n - 1)d = -75 + (n - 1)(12) = 12n - 87

For T(n) < 0, 12n - 87 < 0
12n < 87
n < 7.25
n ≤ 7

Therefore, the minimum of S(n) is S(7) = 6(7)² - 81(7) = -273.
The corresponding value of n is 7.

2014-12-28 00:51:05 補充：
Method 2
Consider completing the square.
S(n)
= 6n² - 81n
= 6[n² - (27/2)n]
= 6[n² - 2(27/4)n]
= 6[n² - 2(27/4)n + (27/4)² - (27/4)²]
= 6[(n - 27/4)² - 729/16]
= 6(n - 27/4)² - 2187/8

2014-12-28 00:51:18 補充：
Therefore, the minimum value of S(n) is -2187/8.
The corresponding value of n is 27/4 = 6.75.
However, n can only take value as positive integer, so this minimum value is not achievable.

Consider
S(6) = 6(6)² - 81(6) = -270
S(7) = 6(7)² - 81(7) = -273

2014-12-28 00:51:32 補充：
Therefore, the true minimum value of S(n) is -273 and the corresponding value of n is 7.