document.write( "Question 1065636: A.p=1,2,3,4........49
\n" ); document.write( "Find a number in a.p such that sum of numbers before it is equal to sum of numbers after it....
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Algebra.Com's Answer #680825 by rothauserc(4718)\"\" \"About 
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the sum of an arithmetic progression is
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\n" ); document.write( "S(n) = n * (a(1) + a(n)) / 2
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\n" ); document.write( "we are given the following
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\n" ); document.write( "n * (n + 1) / 2 = (49 - n - 1) * (n + 2 + 49) / 2
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\n" ); document.write( "multiply both sides of = by 2 and simplify
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\n" ); document.write( "n^2 + n = (48 - n) * (n + 51)
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\n" ); document.write( "n^2 + n = -n^2 - 3n + 2448
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\n" ); document.write( "2n^2 + 4n - 2448 = 0
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\n" ); document.write( "divide both sides of = by 2
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\n" ); document.write( "n^2 + 2n - 1224 = 0
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\n" ); document.write( "use quadratic formula
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\n" ); document.write( "n = (-2 + square root(2^2 + 4 * 1224)) / 2 = 34
\n" ); document.write( "n = (-2 - square root(2^2 + 4 * 1224)) / 2 = -36
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\n" ); document.write( "we reject the negative solution and accept n = 34
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\n" ); document.write( "p = 35
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\n" ); document.write( "check our answer
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\n" ); document.write( "First sum = 34 * (1 + 34) / 2 = 595
\n" ); document.write( "Second sum = 14 * (36 + 49) / 2 = 595
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\n" ); document.write( "our answer checks
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