SOLUTION: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 .... .... .... .... .... ..... a) what's the 3rd number from left in 10th row? b) wha

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Question 979025: 1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
.... .... .... .... .... .....
a) what's the 3rd number from left in 10th row?
b) what's the position of 2009?

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Each row has one more number than the row above.
It is what math teachers call an arithmetic sequence, or an arithmetic progression.
(It depends on where you go to school).
Row n has a%5Bn%5D=n numbers,
and rows 1 through n contain a total of
S%5Bn%5D=sum%28a%5Bi%5D%2Ci=1%2Cn%29=n%28n%2B1%29%2F2 numbers.
That means that the last number on row n is S%5Bn%5D=n%28n%2B1%29%2F2 .
The last number on row 9 is S%5B9%5D=9%2A10%2F2=45 ,
and the third number in row 10 is 45%2B3=highlight%2848%29 .
If we say that number 2009 is in row n%2B1 ,
the last number on row n is n%28n%2B1%29%2F2%3C2009
n%28n%2B1%29%2F2%3C2009
%28n%5E2%2Bn%29%2F2%3C2009
n%5E2%2Bn%3C2%2A2009
n%5E2%2Bn%3C4018
There is more than one way to solve that inequality.
You could find the solutions to the quadratic equation
n%5E2%2Bn=4018<-->n%5E2%2Bn-4018=0
as n=%28-1+%2B-+sqrt%2816073%29%29%2F2 ,
and realize that the solution to n%5E2%2Bn%3C4018<--->n%5E2%2Bn-4018%3C0
n%3C%28-1%2Bsqrt%2816073%29%29%2F2=about+62.9 .
So n=62 marks the row above number 2009 .
The last number on row n=62 is S%5B62%5D=62%2A63%2F2=1953 ,
and you need to go 2009-1953=56 numbers further,
on row n%2B1=62%2B1=63 to get to 2009 .
So, 2009 is the highlight%2856th%29 number on row highlight%2863%29 .