SOLUTION: if S10 = 50, S50 = 0 then what is a10, S100 (Arithmetic Sequence)

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Question 1092219: if S10 = 50, S50 = 0 then what is a10, S100 (Arithmetic Sequence)
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!

We have an arithmetic sequence with
S(10) = 50
S(50) = 0

Since the sum of the first 10 terms is 50 and the sum of the first 50 terms is 0, we know that the sum of terms 11 through 50 must be -50:

A(11)+A(12)+...+A(50) = -50

Now each of the terms 11 through 20 is 10 common differences greater than the corresponding terms 1 through 10, so if the common difference is d,
A(11)+A(12)+...+A(20) = 50+100d

Similarly,
A(21)+A(22)+...+A(30) = 50+200d
A(31)+A(32)+...+A(40) = 50+300d
A(41)+A(42)+...+A(50) = 50+400d

Adding all these together, we have
A(11)+A(12)+...+a(50) = 200+1000d

But we know this sum is -50, so
200%2B1000d+=+-50
1000d+=+-250
d+=+-1%2F4

So the common difference in the sequence is -1/4.

The sum of the first 10 terms is
A(1)+(A(1)+d)+(A(1)+2d)+...+(A(1)+9d) = 10A(1)+45d

The sum of the first 10 terms is 50, so
10A%281%29%2B45d+=+10A%281%29%2B45%28-1%2F4%29+=+50
10A%281%29-45%2F4+=+200%2F4
10A%281%29+=+245%2F4
A%281%29+=+245%2F40+=+49%2F8

So now we have the first term (49/8) and the common difference (-1/4).

We are asked to find the value of A(10), which is A(1)+9d:
A%2810%29+=+49%2F8+%2B+9%28-1%2F4%29+=+49%2F8+-+18%2F8+=+31%2F8

The 10th term is 31/8.

We are also asked to find S(100). S(100) is 100 times the average of A(1) and A(100). A(100) is the first term plus 99 times the common difference.

A%28100%29+=+49%2F8+%2B+99%28-1%2F4%29+=+49%2F8+-+198%2F8+=+-149%2F8
S%28100%29+=+100%28%2849%2F8+-+149%2F8%29%2F2%29+=+100%28-100%2F16%29+=+-625

The sum of the first 100 terms is -625.