I made a program that made the next 25is sequences after 16. It starts at #3 because 30 is #3 Here it is:
30
48
70
96
126
160
198
240
286
336
390
448
510
576
646
720
798
880
966
1056
1150
1248
1350
1456
1566
1680
Looks like 57: 12+9=21, +9=30, +9=39, +9=48, +9=57.
If 3 is the first term, then the nth term is [ 3 x 2(n-1) ] .
If 3 is the first term, then the nth term is [ 3 x 2(n-1) ] .
Sn = 3n2 + 2n - 8
We note the sequence goes up in steps of '8' Hence '8n'. Next for step #1 4#1 ; n = 1 ; 8(1) + c = 16 8 + c = 16 c = 8 Hence the nth terms is 8n + 8 Verifications When n = 3 ; 8(3) + 8 = 24 + 8 = 32 ( which is true).
Looks like 57: 12+9=21, +9=30, +9=39, +9=48, +9=57.
If you mean: 8 28 48 and 68 then the nth term is 20n -12 and so the next number will be 88
> since the value rises by nine at each step and the first term is 12 the formula for > the nth term is: 12+(n-1)*9 Which simplifies to Sn = 9n + 3
If 3 is the first term, then the nth term is [ 3 x 2(n-1) ] .
If 3 is the first term, then the nth term is [ 3 x 2(n-1) ] .
Sn = 3n2 + 2n - 8
We note the sequence goes up in steps of '8' Hence '8n'. Next for step #1 4#1 ; n = 1 ; 8(1) + c = 16 8 + c = 16 c = 8 Hence the nth terms is 8n + 8 Verifications When n = 3 ; 8(3) + 8 = 24 + 8 = 32 ( which is true).
48/30 = 16/10 = 1.6/1 = 1.6
Had it not been for 64, the nth term would be given by t(n) = 3*n2 However, this cannot be the case because then t(4) would have been 48, not 64.
t(n) = 3*2n-1 Or just 3*2n
1.6% of 48= 1.6% * 48= 0.016 * 48= 0.768
48 and 16/30