Formula for the series 8 4 2 1 1/2....please

[manohar]

Can anyone tell me the formula for the series 8,4,2,1,1/2,1/4. I know that the previous number is getting devided by 2, (or multiplied by 0.5)

 

[rlm_hghs03]

it is x divided by two where x is a real number.

x/2

 

[manohar]

Well, let me explain what I am looking for.

For example, the series -4, 6, 16, 26, 36, 46.....is basing ont he formula: [(n * 10) - 14]

1 -4
2 6
3 16
4 26
5 36
6 46
7 56

May I know the similar formula/rule for the series 8, 4, 2, 1, 1/2, 1/4, 1/8........

 

[Denis]

For example, the series -4, 6, 16, 26, 36, 46.....is basing ont he formula: [(n * 10) - 14]
1 -4
2 6
3 16
4 26
5 36
6 46
7 56
May I know the similar formula/rule for the series 8, 4, 2, 1, 1/2, 1/4, 1/8........

Different animals...1st one is an arithmetic series, 2nd one is a geometric series.
Look 'em up with Google.

 

[Deleted member 4993]

If you start with number "a" and get the subsequent numbers by multipling the preceding number by "r" - then the sequence looks like:

ar[sup:3li0e482]0[/sup:3li0e482], ar[sup:3li0e482]1[/sup:3li0e482], ar[sup:3li0e482]2[/sup:3li0e482], ar[sup:3li0e482]3[/sup:3li0e482], ar[sup:3li0e482]4[/sup:3li0e482], .......

so the 1st number is = a*r[sup:3li0e482](1-1)[/sup:3li0e482]

so the 2nd number is = a*r[sup:3li0e482](2-1)[/sup:3li0e482]

so the 3rd number is = a*r[sup:3li0e482](3-1)[/sup:3li0e482]

so the 4th number is = a*r[sup:3li0e482](4-1)[/sup:3li0e482]

so the nth number is = a*r[sup:3li0e482](???)[/sup:3li0e482]