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In an increasing sequence of 10 consecutive integers, the average (ari

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In an increasing sequence of 10 consecutive integers, the average (ari [#permalink]

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28 Dec 2017, 20:52
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In an increasing sequence of 10 consecutive integers, the average (arithmetic mean) of the first five integers is 56. What the average of the last 5 integers in the sequence?

A. 61
B. 58.5
C. 58
D. 57
E. 56.5
[Reveal] Spoiler: OA

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Re: In an increasing sequence of 10 consecutive integers, the average (ari [#permalink]

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29 Dec 2017, 05:02
Bunuel wrote:
In an increasing sequence of 10 consecutive integers, the average (arithmetic mean) of the first five integers is 56. What the average of the last 5 integers in the sequence?

A. 61
B. 58.5
C. 58
D. 57
E. 56.5

Lets consider the first 10 positive integers
1,2,3,4,5 6,7,8,9,10
Grp1 (1-5) has avg of 3 while Grp 2(6-10) has average of 8 => difference of 5
similarly
54,55,56,57,58 59,60,61,62,63
Avg = 56 Avg = 61 Difference is 5

A
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In an increasing sequence of 10 consecutive integers, the average (ari [#permalink]

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29 Dec 2017, 06:21
1
KUDOS
Bunuel wrote:
In an increasing sequence of 10 consecutive integers, the average (arithmetic mean) of the first five integers is 56. What the average of the last 5 integers in the sequence?

A. 61
B. 58.5
C. 58
D. 57
E. 56.5

Takeways from this problem
The average of even consecutive integers is the average of middle two terms(of the sequence)
The average of odd consecutive integers is the middle term of the sequence

Since, the average of the first five terms is 56, the third term of the sequence is 56.
The average of the last five terms, is the eight term of the overall sequence.
Since the sequence is of consecutive integers, the average is 5 more than the average
of the first five numbers, which is 56+5 = 61(Option A)
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In an increasing sequence of 10 consecutive integers, the average (ari [#permalink]

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29 Dec 2017, 07:54
1
KUDOS
Bunuel wrote:
In an increasing sequence of 10 consecutive integers, the average (arithmetic mean) of the first five integers is 56. What the average of the last 5 integers in the sequence?

A. 61
B. 58.5
C. 58
D. 57
E. 56.5

I. Mean = median (very fast method)

In an evenly spaced sequence, the median equals the mean.

56 is the mean of five numbers.
It is also the median. If 56 is the median, there are two integers to its left, and two integers to its right.

The numbers are
54, 55, 56, 57, 58

Last five: 59, 60, 61, 62, 63

Average of the last five integers?
Mean = median
59, 60, 61, 62, 62

II. Average = First + Last divided by 2

If an evenly spaced set, average (arithmetic mean) = $$\frac{FirstTerm+LastTerm}{2}$$

Given an average in a set of consecutive integers, use the formula to find the first term

The first five integers are
x, x+1, x+2, x+3, x+4
Given average: 56
First term: x
Last term (for these five): x+4

$$\frac{x + (x+4)}{2} = 56$$
$$2x + 4 = 112$$
$$2x = 108$$
$$x = 54 =$$ first term of the whole sequence

Average of the last five integers?

Last five integers are
x+5, x+6, x+7, x+8, x+9
(Just pick up where you left off above for the first five)

x = 54. Careful. The first term in this set of five is not x. It is x+5.

First term in last five is (54 + 5) = 59
Last term here is (x+9) = (54 + 9) = 63

Average = $$\frac{FirstTerm+LastTerm}{2}$$

Average = $$\frac{(59+63)}{2} =\frac{122}{2} = 61$$

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In an increasing sequence of 10 consecutive integers, the average (ari   [#permalink] 29 Dec 2017, 07:54
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