Averaging numbers

I have a long colum of numbers,in row A. I would like to average every 4
numbers such as 1-4, 2-6, 3-7, 4-8 etc. and have the answer show in row B
next to the last averaged number. Can someone help me..

JE
Sat Dec 18 08:42:28 CST 2004

One way:

B1: =IF(MOD(ROW(),4)=0,AVERAGE(OFFSET(A1,-3,0,4,1)),"")

Copy down as far as needed.



In article <0aXwd.195922$jE2.174024@bignews4.bellsouth.net>,
"Donduk" <fastcar101@aol.com> wrote:

> I have a long colum of numbers,in row A. I would like to average every 4
> numbers such as 1-4, 2-6, 3-7, 4-8 etc. and have the answer show in row B
> next to the last averaged number. Can someone help me..

Bernd
Sun Dec 19 00:20:21 CST 2004

=SUM(INDEX(A1:A4,0):INDEX(A1:A4,0))/4

into cell B4. Is not volatile and works if you move rows.

HTH,
Bernd

Ragdyer
Sun Dec 19 10:06:01 CST 2004

Isn't this the same thing:

=SUM(A1:A4)/4

In B4, and drag down to copy?
--
Regards,

RD

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"Bernd Plumhoff" <bplumhoff@t-online.de> wrote in message
news:cq36jf$1pv$04$1@news.t-online.com...
> =SUM(INDEX(A1:A4,0):INDEX(A1:A4,0))/4
>
> into cell B4. Is not volatile and works if you move rows.
>
> HTH,
> Bernd
>
>

Bernd
Sun Dec 19 17:10:38 CST 2004

Hi Ragdyer,

You are right. If we want to be independent to insertion of rows (which
means if we want to have an average of this row and the 3 previous rows in
column A) we need =SUM(INDEX(A:A,ROW()-3):INDEX(A:A;A4))/4 I think :-)

Regards,
Bernd

Ragdyer
Sun Dec 19 18:18:51 CST 2004

I'm sorry, but perhaps today I'm a little more dense then I usually am.
I don't see your point in the necessity of using Index().
Besides, I couldn't get your formula to work.

How about this one:

=AVERAGE(INDIRECT("A"&ROW()-3&":A"&ROW()))
--
Regards,

RD

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"Bernd Plumhoff" <bplumhoff@t-online.de> wrote in message
news:cq51po$ceg$03$1@news.t-online.com...
> Hi Ragdyer,
>
> You are right. If we want to be independent to insertion of rows (which
> means if we want to have an average of this row and the 3 previous rows in
> column A) we need =SUM(INDEX(A:A,ROW()-3):INDEX(A:A;A4))/4 I think :-)
>
> Regards,
> Bernd
>
>