1。 1, because S5×S6+ 15=0.

S5=5 so 5 ×S6+ 15=0.

So S6=-3 a6=S6-S5=-8.

Because arithmetic progression Sn=(a 1+an)*n/2.

So s6=(a 1+a6)*3, so a 1=7 d=-3 is deduced.

1。 2. You 'an is the first arithmetic progression with a 1 and a tolerance of d.

Therefore, an = (n-1) * d+a1sn = (a1+an) * n/2 = (2a1+(n-1) d) * n/.

And because S5×S6+ 15=0.

So (5 * a1+10d) * (6 * a1+15d)+15 = 0.

(a 1+2d)*(2a 1+5d)+ 1 = 0

2*a 1？ + 9*a 1*d + 10d？ + 1 = 0

As an equation about a 1, △ ≥ 0.

(9d)？ - 8( 10d？ + 1) ≥ 0

d？ ≥ 8

Get d ≤ -2√2 or d ≥ 2√2.

Because the relationship between a 1 and 0 is uncertain, there are two possibilities for the value of d to be greater than or less than 0.

2. Because S9=S 17, 9a1+36d =17a1+136d.

So d =-0.08a 1

Therefore, this series is the first monotonic decreasing term greater than 0.

Sequence, with a maximum sum sn.

Assuming that an=0 exists, an=a 1+(n- 1)d=0.

So n= 13.5 is not an integer.

So there is no an=0.

When n

Therefore, for sn = a 1+a2+...+ An.

In the case of an > 0, when an starts, sn increases monotonically.

Therefore, when n= 13, the maximum value of sn is the value of s 13.

s 13 = 13 * a 1+78 * d = 13 * a 1-78 * 0.08 a 1 = 6.76 a 1