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chetan2u answer is correct as the number of terms in sequence doesn't matter as
the average in a sequence is a sum of 2 extreme numbers of the sequence.
And in a sequence,mean and median are equal.
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chetan2u answer is correct as the number of terms in sequence can't be even because if the number of terms is even in a sequence, then the average is a sum of 2 middle numbers of the sequence. Since numbers in the sequence are 2.5 more than the previous number, we will not get an average of -10.
So the number of terms is odd.

CaptVarun

The sequence couldn’t be -11.25, -8.75?

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Isn't this true only when the number of integers in the sequence are ODD?
chetan2u
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After the first number in the sequence, each number in sequence T is 2.5 more than the previous number. If the average (arithmetic mean) of the set is -10, what is the median of the sequence?

(A) -25
(B) -10
(C) -8.5
(D) -4
(E) There is not enough information to answer the question.

Let the first number be X, then the sequence is x, x+2.5, x+5, ...

So, T is an Arithmetic Progression with common difference as 2.5, and in an AP, the median and average are always equal, so median=average=-10
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pratiksha1998
Isn't this true only when the number of integers in the sequence are ODD?
chetan2u
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After the first number in the sequence, each number in sequence T is 2.5 more than the previous number. If the average (arithmetic mean) of the set is -10, what is the median of the sequence?

(A) -25
(B) -10
(C) -8.5
(D) -4
(E) There is not enough information to answer the question.

Let the first number be X, then the sequence is x, x+2.5, x+5, ...

So, T is an Arithmetic Progression with common difference as 2.5, and in an AP, the median and average are always equal, so median=average=-10
No, if the set is evenly spaced, the mean will equal the median, regardless of whether the set has an even or odd number of elements.
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