它们的前n项和的比sn/sn'=(n+3)/(n+1),n=10,sn/sn'=(n+3)/(n+1)=13/11
a(n)=a1+(n-1)d aSn=na1+n*(n-1)da/2 =10a1+45d a
b(n)=b1+(n-1)dbSn=nb1+n*(n-1)db/2 =10b1+45d b
(10a1+45d a)/(10b1+45d b)=13/11=(2a1+9da) /(2b1+9db)
a10=a1+9dab10=b1+9dba10/b10=(a1+9da) /(b1+9db)