<PRE><B>Question 1012250</B>
for an arithmetic progression, the nth term is given by
An = A1 + d(n-1) where d is the common difference
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we have two equations in two unknowns
1) A3 = 5 = A1 + d(3-1)
A1 + 2d = 5
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2) A5 = 11 = A1 + d(5-1)
A1 + 4d = 11
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now subtract equation 1 from equation 2 and we get
2d = 6
d = 3
use equation 1 to find A1
A1 + 3(2) = 5
A1 = -1
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equation for the nth term is
An = -1 + 3(n-1)
We can check this by using equation 2
-1 + 3(5-1) = 11
-1 + 12 = 11
11 = 11
our equation checks out
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sum of the first n terms(Sn) = (n(A1 + An)) / 2
A1 = -1
A100 = -1 + 100(100-1) = 9899
S100 = (100(-1 + 9899)) / 2 = 494900

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