Question 1059569
.
The sum of the first n counting/natural numbers is 23005. How many numbers were added (i.e. find n)?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
The sum of the first n counting/natural numbers is {{{S[n]}}} = {{{(n*(n+1))/2}}}.


See the lessons 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Arithmetic-progressions.lesson>Arithmetic progressions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/The-proofs-of-the-formulas-for-arithmetic-progressions.lesson>The proofs of the formulas for arithmetic progressions</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Problems-on-arithmetic-progressions.lesson>Problems on arithmetic progressions</A>,  

in this site.


Therefore, to answer the question, you need to solve this equation


{{{(n*(n+1))/2}}} = 23005,   or


n*(n+1) = 46010.


Apply any method you know to get the answer n = 214.


(Also notice that {{{sqrt(46010)}}} = 214.499. . . .
 It is how I got the answer).
</pre>