watch out for

1 、( 1)d =(a3-a 1)/(3- 1)=(-3- 1)/(3- 1)=-2，

Therefore, an = a1+(n-1) d =1+(n-1) * (-2) = 3-2n.

(2)Sk = k(a 1+AK)/2 = k( 1+3-2k)/2 = k(2-k)=-35，

So k 2-2k-35 = 0, (k+5)(k-7) = 0,

So k = 7 (minus 5).

2. The general term an = a1+(n-1) d = 3-2n,

S (k+2)-sk = a (k+1)+a (k+2) = [3-2 (k+1)]+[3-2 (k+2)] =-4k =-24 (suspected input error).

So k = 6.

3. (1) d = (a6-a3)/(6-3) = (7-4)/(6-3) =1,so an = a3+(n-3) d = 4+(n-3) = n+.

(2)bn = 1/[a(3n+ 1)* a(3n+4)]= 1/[(3n+2)(3n+5)]= 1/3 *[ 1/(3n+2)- 1/(3n+5)]，

So TN =1/3 * [1/5-1/8+11+...+1(.

= 1/3 *[ 1/5- 1/(3n+5)]

= n/[3(3n+5)].

1, (1) d = (a3-a1)/(3-1) = (-3-1)/(3-1) =-2, (2) sk = k (a1+AK)/2 = k (1+3-2k)/2 = k (2-k+5) (k- 35, so k 2-2k-35 = 0, (k+5) (k 2. the general term an = a1+(n-1) d = 3-2n, s (k+2)-sk = a (k+1)+a (k+2) = [3-2 (k+66). 3. (1) d = (a6-a3)/(6-3) = (7-4)/(6-3) =1,so an = a3+(n-3) d = 4+(n-3) = n+. (2) bn =1/[a (3n+1) * a (3n+4)] =1[(3n+2) (3n+5)] =1/3 * [/kloc]