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Hey there,

Is there anything wrong with using S=An? According to this formula, we know that sum will always have to be divisible by 6, so look to see which number isn't divisible by 3 (since they're all even?).

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kilgo16
Hey there,

Is there anything wrong with using S=An? According to this formula, we know that sum will always have to be divisible by 6, so look to see which number isn't divisible by 3 (since they're all even?).

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There is nothing wrong with using this formula.

Sum = Average * Number of terms

But note that you must first figure out that the average will be an integer so sum will be divisible by the number of terms i.e. 6.

Since there will be 6 numbers with a range of 10, the numbers will be like 1, 3, 5, 7, 9, 11 - i.e. alternate numbers. So average will be the number in the middle of 3rd and 4th numbers which will be an integer. Hence the sum will be divisible by 6.
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"I don't understand the term "with a range of 10"..... can anyone plz help me to figure out that?
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"I don't understand the term "with a range of 10"..... can anyone plz help me to figure out that?

In Statistics, the "range" of a set is equal to the largest number in the set minus the smallest number in the set. So in this list, for example:

11, 23, 31, 73

the range is equal to 73 - 11, so the range is 62.

In the question in this thread, the range of the set is 10, so if S is the smallest number in the set, S + 10 will be the largest number.

Hope that helps!
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I used sum of AP here,
We know the formula for sum of terms in AP as SUM=\(n/2(2a+(n−1)d)\)


here n is number of terms i.e 6 here, a is initial number , d is common difference between terms.
SUM =\(6/2(2a+5d)\)

since 2a+5d has to be integer, so options must be divisible by 3. and only option D is not divisible by 3. Hence D is the answer
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