SOLUTION: A.p=1,2,3,4........49 Find a number in a.p such that sum of numbers before it is equal to sum of numbers after it.... Please answer fast...

Algebra ->  Sequences-and-series -> SOLUTION: A.p=1,2,3,4........49 Find a number in a.p such that sum of numbers before it is equal to sum of numbers after it.... Please answer fast...      Log On


   

Question 1065636: A.p=1,2,3,4........49
Find a number in a.p such that sum of numbers before it is equal to sum of numbers after it....
Please answer fast...
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
the sum of an arithmetic progression is
:
S(n) = n * (a(1) + a(n)) / 2
:
we are given the following
:
n * (n + 1) / 2 = (49 - n - 1) * (n + 2 + 49) / 2
:
multiply both sides of = by 2 and simplify
:
n^2 + n = (48 - n) * (n + 51)
:
n^2 + n = -n^2 - 3n + 2448
:
2n^2 + 4n - 2448 = 0
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divide both sides of = by 2
:
n^2 + 2n - 1224 = 0
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use quadratic formula
:
n = (-2 + square root(2^2 + 4 * 1224)) / 2 = 34
n = (-2 - square root(2^2 + 4 * 1224)) / 2 = -36
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we reject the negative solution and accept n = 34
:
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p = 35
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check our answer
:
First sum = 34 * (1 + 34) / 2 = 595
Second sum = 14 * (36 + 49) / 2 = 595
:
our answer checks
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