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 2If 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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