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In a list of K consecutive integers, the median is 240. If K is an odd

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Bunuel
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In a list of K consecutive integers, the median is 240. If K is an odd [#permalink] New post  28 Mar 2018, 03:07
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In a list of K consecutive integers, the median is 240. If K is an odd number, what is the greatest integer in the list?

(A) 239 + (K - 1)/2
(B) 240 + (K - 1)/2
(C) 239 + K/2
(D) 240 + (K + 1)/2
(E) 120 + K
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Re: In a list of K consecutive integers, the median is 240. If K is an odd [#permalink] New post  29 Mar 2018, 17:25
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Bunuel wrote:
In a list of K consecutive integers, the median is 240. If K is an odd number, what is the greatest integer in the list?

(A) 239 + (K - 1)/2
(B) 240 + (K - 1)/2
(C) 239 + K/2
(D) 240 + (K + 1)/2
(E) 120 + K


We can let K = 3, so the list of numbers would be 239, 240, 241.

Looking at the answer choices we see that B must be correct since 240 + (3 - 1)/2= 241.

Answer: B
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In a list of K consecutive integers, the median is 240. If K is an odd [#permalink] New post  28 Mar 2018, 04:00
In an AP series , the median or middle term is given by = (first term + last term) /2
Let the last term =L
first term = A
In an AP , last term is given by = a+[n-1]d , where n= no of terms and d= common difference
Here n=k, and d=1 (consecutive numbers)
hence ,
240= [a+a+(k-1)]/2
this gives a= 240- (k-1)/2
hence the largest integer or last term = a+ (k-1)
substituting a, we get
last term or largest term = 240+ (k-1)/2
option B
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Re: In a list of K consecutive integers, the median is 240. If K is an odd [#permalink] New post  28 Mar 2018, 13:58
We are given that a list contains K consecutive integers and that K is odd. So the median term is the \(\frac{(K+1)}{2}\)th term of \(K\) terms. Which means that there are \(K - \frac{(K+1)}{2}\) terms from the median to the max term.

(median) + (# of terms to get to the max) = max term
\(240 + K - \frac{(K+1)}{2}\)

\(240+\frac{2K-K-1}{2}\)

\(240+\frac{K-1}{2}\)

Answer: B
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In a list of K consecutive integers, the median is 240. If K is an odd [#permalink] New post  28 Mar 2018, 20:27
Bunuel wrote:
In a list of K consecutive integers, the median is 240. If K is an odd number, what is the greatest integer in the list?

(A) 239 + (K - 1)/2
(B) 240 + (K - 1)/2
(C) 239 + K/2
(D) 240 + (K + 1)/2
(E) 120 + K


let K=3
greatest term=241
241=240+(3-1)/2
B
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Re: In a list of K consecutive integers, the median is 240. If K is an odd [#permalink] New post  16 Mar 2022, 11:50
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Re: In a list of K consecutive integers, the median is 240. If K is an odd [#permalink]
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