Question 291098
<pre><font size = 4 color = "indigo"><b>
The sum of integers from -25 to +25 is obviously 0. The numbers from -25 
to -1 makes 25 numbers and the numbers from 1 to 25 makes another 25 numbers.
Then the 0 between -1 and +1 makes a 51st number and if we put 26 on the upper
end the sum will be 26.  So there must be 52 numbers in the set.

But if you're supposed to work it out with an algebraic  formula, 
then you'll have to do it this way:

{{{a[1]=-25}}}  {{{d=1}}},  {{{S[n]=26}}}.  Find n

Formula needed:

{{{S[n]=(n/2)(2a[1] + (n-1)d)}}}

{{{26=(n/2)(2*(-25) +(n-1)1)}}}

{{{26=(n/2)(-50 +n-1)}}}

{{{26=(n/2)(-51 +n)}}}

Multiply both sides by 2

{{{52=n(-51+n)}}}

{{{52=-51n+n^2}}}

{{{0=n^2-51n-52}}}

Factor the right side:

{{{0=(n-52)(n+1)}}}

Setting each factor equal to 0

{{{n-52=0}}} gives {{{n=52}}}

{{{n+1=0}}} gives {{{n=-1}}}

We cannot have a negative number of terms, so

There are 52 terms, or numbers in the set.

Edwin</pre>