IlCreatore wrote:
We can proceed by steps:
1) number of elements in the two sets: we have to take the first and the last multiple, then compute last - first, divide by the number whose multiples are of interest and sum 1. In our case, in column A, we have 100−44+1=25, while for column B we get 100−55+1=20.
2) Compute the sum. The formula for the sum of an arithmetic progression is sum=n2(first+last), where n is the number of elements in the progression and first and last are the first and the last elements. Thus, for column A, 252(4+100)=1300, while column B equates 202(5+100)=1050.
We conclude that A is greater!
I doubt the above explanation, kindly provide some feed back.
As we need to find the multiples of 4 & 5 less than 100, i.e. 100 exclusiveQTY A :: since the number has to be less than 100
The number of terms =
96−44+1=24 ( last multiple of 4, less than 100 - first multiple of 4 )
The average =
99+12=50Hence the multiples of 4 less than 100 = 50 * 24
QTY B::
The number of terms =
95−55+1=19The average =
99+12=50Hence the multiples of 5 less than 100 = 50 * 19
Therefore QTY A > QTY B