Divide the first record of a column by the last record

Question

I have a table showing data reported during the one month period:

DATA       REGISTRO
01/06/2016  
02/06/2016  
03/06/2016  
04/06/2016  
05/06/2016  50
06/06/2016  60
07/06/2016  10
08/06/2016  20
09/06/2016  50
10/06/2016  60
11/06/2016  100
12/06/2016  50
13/06/2016  10
14/06/2016  30
15/06/2016  10
16/06/2016  50
17/06/2016  10
18/06/2016  30
19/06/2016  30
20/06/2016  40
21/06/2016  50
22/06/2016  30
23/06/2016  10
24/06/2016  50
25/06/2016  10
26/06/2016  30
27/06/2016  30
28/06/2016  40
29/06/2016  50
30/06/2016  60

I need to divide the first record by the last, the problem and that the record can start on any date,:

DATA       REGISTRO
01/06/2016  
02/06/2016  30
03/06/2016  20
04/06/2016  55
05/06/2016  50
06/06/2016  60
07/06/2016  10


DATA       REGISTRO
01/06/2016  
02/06/2016  
03/06/2016  20
04/06/2016  55
05/06/2016  50
06/06/2016  60
07/06/2016  10

Likewise the last. That is, I would need to be able to identify which is the first record and which is the last and divide one by the other.

Answer 1

Otácio, is it to divide the dates (are numbers internally) or the value of the records? In this case the initial values of the records are zero (blank cell).

Regardless of what, assuming you divide the dates, solve so considering cell A2 as the initial):

=INDIRETO("A"&(CONT.SE(A2:A99999;">0")+1))/A2

The CONT.SE takes the amount of items in column A, sum 1 to get the number of the last row, taking the INDIRECT column A and this row number, it will bring the value of the final cell and dividing the result by A2 you get the desired result.

See if that’s it.