Let a1,a2,a3 , be three arithmetic mean between 11 and 12 So, the AP series is 11,a1,a2,a3,12a is the first term and d is the common difference of AP. Then a = 11t5=a+5d⇒12=11+5d⇒d=51∴a1=11+d=11+51=556a2=11+2d=11+52=557a3=11+3d=11+53=558Hence, the required arithmetic means are 556,557,558