等差数列Sn=(a1+an)*n/2
∴S10=(a1+a10)*10/2=30
a1+a10=6
a1+(a1+9d)=6
2a1+9d=6 ①
S30=(a1+a30)*30/2=10
a1+a30=2/3
a1+(a1+29d)=2/3
2a1+29d=2/3 ②
由 ① ②,得 d=-4/15,a1=21/5
∴S40=(a1+a40)*40/2
=(a1+a1+39d)*20
=(21/5+21/5+39*-4/15)*20
=-40
等差数列Sn=(a1+an)*n/2
∴S10=(a1+a10)*10/2=30
a1+a10=6
a1+(a1+9d)=6
2a1+9d=6 ①
S30=(a1+a30)*30/2=10
a1+a30=2/3
a1+(a1+29d)=2/3
2a1+29d=2/3 ②
由 ① ②,得 d=-4/15,a1=21/5
∴S40=(a1+a40)*40/2
=(a1+a1+39d)*20
=(21/5+21/5+39*-4/15)*20
=-40