Two Arithmetic SeriesDate: 7/1/96 at 7:27:6 From: Novica Subject: Two Series an and bn are two arithmetic series, An and Bn are sums of first n elements. If (4n+27) * An = (7n+1) * Bn, compute a11/b11. Yours truly, Dr. Nele Date: 7/1/96 at 11:56:16 From: Doctor Anthony Subject: Re: Two Series Let the 'a' series be a, (a+d), (a+2d) etc Let the 'b' series be b, (b+e), (b+2e) etc Then with n=1 in the given equation (4+27)a = (7+1)b 31a = 8b and a/b = 8/31 An = (n/2)(2a+(n-1)d) and Bn = (n/2)(2b+(n-1)e) In the series of equations connecting An and Bn we shall omit the term (n/2) as it appears on both sides of the equation. With n=2 we have 35(2a+d) = 15(2b+e) With n=3 we have 39(2a+2d) = 22(2b+2e) We now write out these equations to get: 70a - 30b + 35d - 15e = 0 78a - 44b + 78d - 44e = 0 Dividing through by b, we get for these equations 70(a/b) - 30 + 35(d/b) - 15(e/b) = 0 and 78(a/b) - 44 + 78(d/b) - 44(e/b) = 0 but we know a/b = 8/31 so 70(8/31) - 30 + 35(d/b) - 15(e/b) = 0 and 78(8/31) - 44 + 78(d/b) - 44(e/b) = 0 Collecting and simplifying terms we have: 35(d/b) - 15(e/b) = 370/31 and 78(d/b) - 44(e/b) = 740/31 These equations lead to (d/b) = 14/31 (e/b) = 8/31 Now we can evaluate a11/b11 since we know a/b = 8/31, d/b = 14/31, and e/b = 8/31 a11/b11 = (a+10d)/(b+10e) = (a/b + 10(d/b))/(1 + 10(e/b)) = (8/31 + 140/31)/(1 + 80/31) = (8 + 140)/(31 + 80) = 148/111 = 4/3 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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