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19 Feb 2022, 19:54
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Bunuel, why it is not correct that sum of 21 numbers=21*Average? Why it has to be sum21=sum20+n=5n+n=6nsum21=sum20+n=5n+n=6n;?
20 Feb 2022, 04:05
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tkorzhan1995
The sum is ALWAYS equal to (average)*(number of terms), so \(sum_{21}=average_{21}*21\). But we don'ts know \(average_{21}\), we know only \(average_{20}\), so we have to calculate \(sum_{21}\) in another way: \(sum_{21}=sum_{20}+n=5n+n=6n\).
Hope it helps.
26 Sep 2024, 02:21
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i always prefer to consider the mean of n element as the value that, multiplied by the number n is equal to the sum of the n elements.
in this case
n = 4*AVG(20)
thus
20*AVG(20) = SUM(20)
SUM(21) = SUM(20) + n = 20*AVG(20)+ 4*AVG(20) = 24*AVG(20)
we want the ratio n/sum(21)
n/SUM(21) = (4*AVG(20)) / (24*AVG(20)) = 1/6
n = 1/6 SUM(21)
in this case
n = 4*AVG(20)
thus
20*AVG(20) = SUM(20)
SUM(21) = SUM(20) + n = 20*AVG(20)+ 4*AVG(20) = 24*AVG(20)
we want the ratio n/sum(21)
n/SUM(21) = (4*AVG(20)) / (24*AVG(20)) = 1/6
n = 1/6 SUM(21)
