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 Albert.Teng October 1st, 2012 05:54 AM

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

 Isbell October 1st, 2012 08:40 AM

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

$99=\frac{19+a_2}{2}$

$a_2=2(99)-19=179$

 Albert.Teng October 1st, 2012 09:16 AM

Re: Sequence: find a2

Isbell :perfect :!:

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