Question 297622
The sum of the first 100 terms of the sequence 1, -2, 3, 4, -5, 6, 7, -8, 9, 10… is 1750. The sum of the first 100 terms of the sequence 1, 2, -3, 4, 5, -6, 7, 8, -9, 10… is equal to
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Let the answer be x.

1, -2, 3, 4, -5, 6, 7, -8, 9, 10…

The signs go in groups of three +, -,  +,  so the last group of three
less than 100 that have signs like that will end in a multiple of 3,
which is the group +97, -98, +99, since 99 is the largest multiple of 99.
The last term would be +100 because 100 would be the beginning of the next 
group of three that go +, -, +, so we have:

1 - 2 + 3 + 4 - 5 + 6 + 7 - 8 + 9 + 10 + … + 97 - 98 + 99 +100 = 1750.

Next we look at the second sequence:

1, 2, -3, 4, 5, -6, 7, 8, -9, 10…

The signs also go in groups of three +, +,  -,  so the last group of three
less than 100 that have signs like that are +97, +98, -99, so the last
term would be +100 because it would be the beginning of the next group
of three that go +, +, -, so we have:

1 + 2 - 3 + 4 + 5 - 6 + 7 + 8 - 9 + 10 + … + 97 + 98 - 99 + 100 = x.

Now let's add the two sequences together term by term:

1 - 2 + 3 + 4 - 5 + 6 +  7 - 8 + 9 + 10 + … +  97 - 98 + 99 + 100 =   1750  
1 + 2 - 3 + 4 + 5 - 6 +  7 + 8 - 9 + 10 + … +  97 + 98 - 99 + 100 =      x
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2 + 0 + 0 + 8 + 0 + 0 + 14 + 0 + 0 + 20 + … + 194 +  0 +  0 + 200 = 1750+x

The left side is the arithmetic sequence with {{{d=6}}} and {{{a[1]=2}}} 

2 + 8 + 14 + 20 + ... + 194 + 200

To find the number of terms, n, we can do it by observing the
groups of 3, but if you can't tell there would be {{{99/3+1}}} 
or 34 terms, then we can calculate the n=34 this way:

{{{a[n]=a[1]+(n-1)d}}}
{{{200=2+(n-1)6}}}
{{{198=6(n-1)}}}
{{{33=n-1}}}
{{{34=n}}}

So

{{{S[n]=(n/2)(a[1]+a[n])}}}
{{{S[34]=(34/2)(2+200)}}}
{{{S[100]=(17)(202)=3434}}}

Therefore the equation

2 + 0 + 0 + 8 + 0 + 0 + 14 + 0 + 0 + 20 + … + 194 +  0 +  0 + 200 = 1750+x

becomes

{{{3434=1750+x}}}
{{{1684=x}}}

and the required sum is 1684.

Edwin</pre>