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a, b, c, d, e, f are some numbers such that a < b < c < d < e < f, and
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23 Oct 2024, 22:58
23 Oct 2024, 22:58
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Question Stats:
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66% (01:50) wrong based on 3 sessions
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\(a, b, c, d, e, f\) are some numbers such that \(a < b < c < d < e < f\), and all of the numbers are positive integers. Of the following, which options are true?
A. If the difference between any two successive numbers is same, mean of the above numbers is \(\frac{a+f}{2}\)
B. If option A. above holds ture, the median of the numbers is \(\frac{a+f}{2}\)
C. If a – c = d – f, then the numbers given above are in arithmetic sequence.
A. If the difference between any two successive numbers is same, mean of the above numbers is \(\frac{a+f}{2}\)
B. If option A. above holds ture, the median of the numbers is \(\frac{a+f}{2}\)
C. If a – c = d – f, then the numbers given above are in arithmetic sequence.
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Re: a, b, c, d, e, f are some numbers such that a < b < c < d < e < f, and
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24 Oct 2024, 09:42
24 Oct 2024, 09:42
why is C not true
Re: a, b, c, d, e, f are some numbers such that a < b < c < d < e < f, and
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24 Oct 2024, 10:09
24 Oct 2024, 10:09
1
Expert Reply
OE
We know that a, b, c, d, e & f are positive integers, such that a<b<c<d<e< f; we need to check from the options that which of them is true.
(A). the mean of the above numbers is (a + f)/2, if the difference between any two successive numbers is same — which is true as whenever a set of numbers form arithmetic sequence i.e. they have same gap between any two consecutive numbers the average (mean) comes same as the median which is same as the average of the first & the last term of the set.
(B) the median of the numbers is (a +f)/2, if option A holds true — which is true as the average is same as the median in case of same gap between any two consecutive terms.
(C) if a — c=d-f, then the numbers given above are in arithmetic sequence — which might not be true as the gap between c and d is not known. For example if we take a = 2, b = 4, c = 6, d = 10, e = 12 & f = 14, we get a—c=d—f= -4 but the numbers 2, 4, 6, 10, 12, 14 do not form an arithmetic sequence.
Hence only statements (A) & (B) are true, so are correct.
We know that a, b, c, d, e & f are positive integers, such that a<b<c<d<e< f; we need to check from the options that which of them is true.
(A). the mean of the above numbers is (a + f)/2, if the difference between any two successive numbers is same — which is true as whenever a set of numbers form arithmetic sequence i.e. they have same gap between any two consecutive numbers the average (mean) comes same as the median which is same as the average of the first & the last term of the set.
(B) the median of the numbers is (a +f)/2, if option A holds true — which is true as the average is same as the median in case of same gap between any two consecutive terms.
(C) if a — c=d-f, then the numbers given above are in arithmetic sequence — which might not be true as the gap between c and d is not known. For example if we take a = 2, b = 4, c = 6, d = 10, e = 12 & f = 14, we get a—c=d—f= -4 but the numbers 2, 4, 6, 10, 12, 14 do not form an arithmetic sequence.
Hence only statements (A) & (B) are true, so are correct.
