So a15 = a1+14d = 29+14d > 0, and a16 = a1+15d = 29+/. 0, so d=-2, so the sum of the first 15 items is the largest, and the maximum value is15a1+[(15 *14)/2] * (-2) = 225.

2. According to the meaning of the question: a3>0, a9<0, a3=-a9, that is, a3+a9=2a6=0, so a6=0, that is, the first five items are all positive, the sixth item is 0, and everything from the seventh item is negative, so the sum of the first five items is equal to the sum of the first six items, which is the largest sum of the first N items.

3. Because a3+a7=2a5=-6, a5=-3, and because A1=-1,the tolerance d=(a5-a 1)/4=2, and An =-/kloc.

4. S9 = 9 (a1+a9)/2 = 9a5 =18, so a5=2, sn = n (a1+an)/2 = n (a5+an-4)/2 = n (.