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 \(\frac{100-4}{4}+1 = 25\), while for column B we get \(\frac{100-5}{5}+1 = 20\).
2) Compute the sum. The formula for the sum of an arithmetic progression is \(sum = \frac{n}{2}(first+last)\), 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, \(\frac{25}{2}(4+100) = 1300\), while column B equates \(\frac{20}{2}(5+100) = 1050\).
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 = \(\frac{96-4}{4}+1 = 24\) ( last multiple of 4, less than 100 - first multiple of 4 )
The average = \(\frac{{99 + 1}}{2} = 50\)
Hence the multiples of 4 less than 100 = 50 * 24
QTY B::
The number of terms = \(\frac{95-5}{5}+1 = 19\)
The average = \(\frac{{99 + 1}}{2} = 50\)
Hence the multiples of 5 less than 100 = 50 * 19
Therefore QTY A > QTY B