In the arithmetic sequence 3, 12, 21, 30, 39, 48 912 is the second______ The number of them

In the arithmetic sequence 3, 12, 21, 30, 39, 48 912 is the second______ The number of them


(912-3) △ 9 + 1, = 909 △ 9 + 1, = 101 + 1, = 102; so the answer is: 102



Can you find a rule to fill in? 3,12,21,30,39,48,57… (1) The 12th number is______ (2) 912 is the second______ The number of them


(1) 3 + (12-1) × 9, = 3 + 11 × 9, = 3 + 99, = 102; the 12th item is 102; (2) let the nth item be 912, then 3 + (n-1) × 9 = 912, & nbsp; (n-1) × 9 = 909, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; n-1 = 101, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; n = 102; 912 is item 102



Choose three different numbers from the set {1,2,3,4,..., 21}, and make these three numbers into an arithmetic sequence, which has at most one
Can you help me


Tolerance d = 1:19
D = 2: 17
D = 3:15
D = 4:13
D = 5: 11
D = 6:9
D = 7: 7
D = 8:5
D = 9:3
D = 10:1