since the production increases by a fixed no therefore the production is in AP
production of car in 3rd year=
a+2d=1100 (1)
production of car in 11th year=
a+10d=2700 (2)
subtracting (1) from (2) we get
a+10d-(a+2d)=2700-1100
a+10d-a-2d=1600
8d=1600
d=200
substituing the value od d in equation(1) we get
a+2(200)=1100
a+400=1100
a=1100-400
a=700
production of car in 20th year=
a+19d=a20
700+3800=a20
a20=4500
so the production of car in 20th year is 4500
total car produced in 20 years=
n/2(2a+(n-1)d)=sn
s20=20/2{2(700)+19(200)}
s20=10{1400+3800}
s20=10(4200)
s20=42000
therefore 42000 cars will be produced in 20years