a=20, d=10 in this AP. Initial mean (average of first and last) is (a+a+(N-1)d)/2=(2a+(N-1)d)/2=(40+10N-10)/2=15+5N=50, where N=7, the number of terms in the series. Smallest mean is (20+30)/2=(a+a+10)/2=a+5=25. Largest mean is (a+(N-1)d+a+(N-2)d)/2=(2a+(2N-3)d)/2=(2a+110)/2=a+55=75. 75/25=3.
If we select only multiples of 10 (d=10) we have (20,30), (20,40), (20,50), (20,60),... (6 pairs); (30,40), (30,50),... (5 pairs); (40,50),...(4 pairs); ... (70,80) (1 pair). The smallest mean is (20+30)/2=25; the largest is (70+80)/2=75 and 75:25 is 3:1. The means are not unique (e.g., 20+50=30+40, 20+80=30+70=40+60). If we write down the means for each pair group we have: 25, 30, 35, 40, 45, 50 for 6 pairs; 35, 40, 45, 50, 55 for 5 pairs; 45, 50, 55, 60 for 4 pairs; 55, 60, 65 for 3 pairs; 65, 70 for 2 pairs and 75 for 1 pair. Therefore the distinctly different means are: 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, so n=11.