My Math Forum Sequence: find a2

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 October 1st, 2012, 05:54 AM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Sequence: find a2 $\text a\,\, sequence \,\,a_1,a_2,a_3,--------------,a_n\text$ $\text given :\text$ $(1) \text a_1=19, \,\,a_9=99\text$ $(2)\text for \,\,any \,\,n\geq 3 , \,\, \,\, a_n=\frac {a_1+a_2+a_3+ -------------+a_{n-1}}{n-1}\text$ $\text please \,\, find \,\, a_2\text$ Ans :179
 October 1st, 2012, 08:40 AM #2 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 1 Re: Sequence: find a2 We are given a sequence with $a_1=19$ $a_3=\frac{a_1+a_2}{2}$ $a_4=\frac{a_1+a_2+a_3}{3}=\frac{a_1+a_2}{3}+\frac{ a_3}{3}=\frac{2a_3}{3}+\frac{a_3}{3}=a_3$ $a_5=\frac{a_1+a_2+a_3+a_4}{4}=\frac{a_1+a_2+a_3}{4 }+\frac{a_4}{4}=\frac{3a_4}{4}+\frac{a_4}{4}=a_4=a _3$ $a_6=\frac{a_1+a_2+a_3+a_4+a_5}{5}=\frac{a_1+a_2+a_ 3+a_4}{5}+\frac{a_5}{5}=\frac{4a_5}{5}+\frac{a_5}{ 5}=a_5=a_3$ and so on. So, we are actually dealing with a sequence that looks like this: $19,\; a_2, \; a_3, \; a_3,\; a_3, \; a_3,\; a_3, \; a_3,\; a_3, \; ......,$ Therefore $a_9=a_3=\frac{a_1+a_2}{2}$ or $99=\frac{19+a_2}{2}$ $a_2=2(99)-19=179$
 October 1st, 2012, 09:16 AM #3 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: Sequence: find a2 Isbell erfect

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