an = a 10+(n- 10)2 = 30+2n-20 = 10+2n

Sn = (a1+an) n/2 = (12+10+2n) n/2 = (11+n) n = 242, so n =/kloc.

2.an = a 1+(n- 1)d = 50-0.6n+0.6 = 50.6-0.6n

50.6-0.6n < 0, so n > 253/3 = 84.333, so start from item 85.

sn =(a 1+an)n/2 =( 100.6-0.6n)n/2 = 50.3n-0.3n？

An is less than 0 from the 85th item in the previous step, so it is greater than 0 before the 84th item.

So the maximum sn is s84 =50.3x84-0.3*84? =4225.2-2 1 16.8=2 108.4

3.an=2 1-4(n- 1)=25-4n

an>0 n7

4, odd number and flavor s 1

Even terms sum taste s2

s 1+s2=354

s 1/s2=27/32

s 1=27*6= 162

s2=32*6= 192

s2-s 1=30=6d

d=5