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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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New post 28 Dec 2017, 21:52
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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

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

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New post 29 Dec 2017, 06:02
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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New post 29 Dec 2017, 07:21
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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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New post 29 Dec 2017, 08:54
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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

ANSWER A

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

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

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New post 21 Mar 2018, 03:12
(x + 1) + (x + 2) + (x +3) + (x + 4) + (x + 5) = 280

now (x + 6), (x +7), (x + 8), (x + 9), (x + 10), each is exactly 5 greater than its counterparts, namely (x + 1), (x + 2) and so on ...

thus, there is total of (5 * 5) = 25 greater than 280, so the sum of last 5 numbers is (280 + 25) = 305, and average is (305/5) = 61

also, the problem is solvable if the numbers are plotted accordingly

as the mean, the 3rd number, equals to 56, the series is

54, 55, (56), 57, 58
and last 5 numbers with mean 61 are

59, 60, (61), 62, 63

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

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New post 21 Mar 2018, 18:45
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


if 1st term=x,
then x+2 and x+7 will be the averages of 1st 5 & 2nd 5 terms respectively
x+7-(x+2)=5
56+5=61
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Re: In an increasing sequence of 10 consecutive integers, the average (ari  [#permalink]

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New post 22 Mar 2018, 15:47
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


Solution:

We can let x = 1st integer, and thus x + 4 = 5th integer. Since we have a consecutive set of integers, we can add the 1st integer and the 5th and divide by 2 to get the average; thus:

(x + x + 4)/2 = 56

2x + 4 = 112

2x = 108

x = 54

The first integer is 54, so the 6th integer is 54 + 5 = 59, and the 10th integer is 54 + 9 = 63.

Thus, the average of the last 5 integers is (59 + 63)/2 = 61.

Alternate Solution:

In a increasing sequence of 10 consecutive integers, the 6th integer will be 5 more than the 1st integer, the 7th integer will be 5 more than the 2nd integer, and so on. Since each of the last 5 integers is 5 more than its counterpart in the first 5 integers. The average of the last 5 integers should be 5 more than the average of the first 5 integers also. Since we are given that the average of the first 5 integers is 56, the average of the last 5 integers is 56 + 5 = 61.

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

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New post 26 Mar 2018, 10:05

Solution 1:



Given:

    • An increasing sequence of 10 consecutive integers.

    • Average of first 5 integers = \(56\)

Working out:

We need to find out the average of the last 5 integers.

Since the sequence contains consecutive integers, let us assume that the numbers are:

\(a, a+1, a+2, a+3…. a +9.\)

    • Average of first five = \(56\)

      o Sum of first five numbers = \(56*5 = 280\)

      o Thus, \(a+ (a+1) + (a+2) + (a+3) + (a+4) = 280\)

         Or, 5a + 10 = 280

         Or, \(5a = 270\),

         Or, \(a = 54\).


    • Thus, average of the last 5 numbers = \((a+5) +(a+6)+ (a+7) + (a+8) + (a+9) / 5\)

      o Or, \(5a + 35 / 5\)

      o Or, \(270 + 35 / 5\)

      o Or, \(305 / 5 = 61\).


Thus, the average of the last 5 terms = \(61\).

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

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New post 26 Mar 2018, 10:09

Solution 2:



Given:

    • An increasing sequence of 10 consecutive integers.

    • Average of first 5 integers = 56

Working out:

We need to find out the average of the last 5 integers.

Since the sequence contains consecutive integers, let us assume that the numbers are:

\(a, a+1, a+2, a+3…. a +9.\)

To understand the solution, let us visualize the problem.

Also, keep in mind that the average of consecutive terms is always the middle term when the total number of terms is ODD.

Image

From the first half of the diagram, it is known that the value of \(a = 54\).

Thus, \(a+ 7 = 54 + 7 = 61\), and this is the average of the last 5 numbers.

Answer: Option A
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In an increasing sequence of 10 consecutive integers, the average (ari &nbs [#permalink] 26 Mar 2018, 10:09
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