a)

Each time the number changes to a bigger one, it increases by 2.



1 2
2 4's
3 6's
4 8's
5 10's
6 12's


1+ 2 + 3 + 4 + 5 + 6 = 21
As you can see from your working, the 16th to 21st number are all 12's.

So the 20th number must be 12.


Do you notice another pattern? For each number, the value is always twice the total number of the same kind of numbers.

Eg. There are 3 number 6s
4 number 8s


b)

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78 (using calculator or manual)

Or

Notice that 1+12 = 13
2+11 =13
3 + 10 = 13
.
.
6 + 7 = 13

We can get 6 pairs of 13. The number of pairs is half the last number. The total in each pair is the same, the sum of the first and last number.

6 x 13 = 78

So the number with 12 of the same kind is
12 x 2 = 24

The 78th number is therefore 24

Then, the next 13 numbers are 2 more than 24, which is 26.

So the 79th, 80th,81st, 82nd ... 89th, 90th, 91st are all 26's.

So the 80th number is 26.




c) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

So the last number here, the 45th number, is 9 x 2 = 18

The next 10 numbers are all 2 more than 18, which is 20.

So the 46th, 47th,48th,49th 50th numbers are all 20's. There are 5 of them.


Sum of first 50 numbers

= 1 x 2 + 2 x 4 + 3 x 6 + 4 x 8 + 5 x 10 + 6 x 12 + 7 x 14 + 8 x 16 + 9 x 18 + 5 x 20
= 670

I wonder if there is a way to write the nth term down in a single mathematical formula.

Yes, there are some formulas. You can do a google search and look up the result for the OEIS. There are also some forums discussing the formula and its validity for higher numbers

Thanks I can also learn

Welcome