a1=0.1a0+10=0.1*100+10=20
a2=0.1a1+10=0.1*20+10=12
a3=0.1a2+10=0.1*12+10=11.2
a4=0.1a3+10=0.1*11.2+10=11.12
(a5=11.112, a6=11.1112, etc.)
a[n]=100/9+(8/9)*10^(2-n) for n>0
The nth term approaches 11 1/9=100/9 as n approaches infinity, but a[n]>100/9 so never reaches this value.