What is the 1st number in the 100th term?

Question

The terms are (1) (2,3) (4,5,6) (7,8,9,10) (11,12,13,14,15)
pls show you arrived at this answer and include any necessary formulas that you used to derive the answer

yes, it's a math league question

Answer 1

just out of curiosity, was this from math league?

ok the solution:
first "term": 1
second "term": (2,3) 3=1+2
third "term": (4,5,6) 6=1+2+3
fourth "term": (7,8,9,10) 10=1+2+3+4
...
so you see
the last number of the 100th term can be found by adding 1+2+3+4+5+6+7+...+97+98+99+100
the formula for finding the sum of an arithmetic sequence (a sequence of numbers with the same difference between each term) is:
((first term)+(last term))*(number of terms) / 2
thus (1+100)*100/2=5050
so 5050 is the last number of the 100th term
the first number of the 100th term would be
5050-100+1=4951

Answer 2

The 1st term has 1 number
The 2nd term has 2 numbers
.
The nth term has n numbers

The first n terms have n(n+1)/2 numbers
The first 99 terms have 99*100/2 = 4950 numbers

The first number in the 100th term will be 4950+1 = 4951

Answer 3

If the first term contains one number, and the second term contains two numbers, the nth term contains n numbers. Therefore, the total number of numbers in the first n terms is:
Sum[i, i=1, i=n] = 1/2*n*(n+1)

For the first 99 terms, there are
1/2*99*(99-1) = 4950 numbers

Since the numbers just follow the integers, the first number in the 100th term is "4951".

Answer 4

The 1st term has 1 number
The 2nd term has 2 numbers
.
The nth term has n numbers

The first n terms have n(n+1)/2 numbers
The first 99 terms have 99*100/2 = 4950 numbers

The first number in the 100th term will be 4950+1 = 4951

Answer 5

1+2+3+...+ n = n(n+1)/2
so the n-term will start with

(n-1)n/2+1 = (n^2-n+2)/2

therefore the 100-term will start with

99*100/2 +1 = 99* 50 + 1 =4951