Around the World in 80 Questions (Day 10): In a set of k consecutive

Question

In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A.  \frac{21k}{2}

B.  \frac{21 - k}{2}

C.  \frac{41 - k}{2}

D.  \frac{42 - k}{2}

E.  \frac{43 - k}{2}

Kinshook · Accepted Answer

Given: In a set of k consecutive integers, where k is two-digit number, the largest number is 21. 
Asked: What is the average (arithmetic mean) of the set in terms of k ?

First term a = 21
common difference = -1
Last term l = 21 + (k-1)(-1) = 21 + 1-k = 22 -k

Arithmetic mean = Sum / k = k/2(a+l)/k = (21 + 22 - k)/2 = (43-k)/2

IMO E

nivivacious · Answer

The average of a set of consecutive integers is the middle number if the count of numbers (k) is odd, or the average of the two middle numbers if k is even.
In this case, the largest number is 21 and there are k consecutive numbers. So, the smallest number is 21 - k + 1.
The average of the set would be (smallest number + largest number) / 2 = (21 + 21 - k + 1) / 2 = (42 - k + 1) / 2 = (43 - k) / 2.
So, the correct answer is E. (43−k)/2.

limuengaoey · Answer

In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?
suppose k is 3 hence (19+20+21)/3 = 20
try to plug in 
a = 21*3/2 =31.5
b= (21-3)/2 = 9
c=(41-3)/2 = 19
d = (42-3)/2=19.5
e=(43-3)/2 = 20
the answer is E

ankitgoswami · Answer

Largest term =  21

Total k terms, so first term =  21 - (k-1)

So average =  \frac{{21-(k-1)+21}}{2}

=  \frac{43 - k}{2}

Ans  E

missionmba2025 · Answer

Assume k = 11

Smallest Value = 21 - 11 + 1 = 11

11 12 13 14 15  16  17 18 19 20 21

In consecutive numbers --> mean = median. In this example mean = median = 16.

A.  \frac{21k}{2}  = (21*11)/2 = Not 16 : Eliminate.

B.  \frac{21 - k}{2} = (21-11)/2 = Not 16 : Eliminate.

C.  \frac{41 - k}{2}  = (41-11)/2 = Not 16 : Eliminate.

D.  \frac{42 - k}{2}  = (42-11)/2 = Not 16 : Eliminate.

E.  \frac{43 - k}{2} = (43-11)/2 = 32/2 = 16 : Our Answer.

IMO E

PK1 · Answer

In a set of k consecutive integers, where k is two-digit number, the largest number is 21. What is the average (arithmetic mean) of the set in terms of k ?

A.  \frac{21k}{2}

B.  \frac{21 - k}{2}

C.  \frac{41 - k}{2}

D.  \frac{42 - k}{2}

E.  \frac{43 - k}{2}

Answer:
K consecutive integers
 Largest number = 21
so 1st number = 21- K +1= 22 - K
Sum of K consecutive numbers. = K/2(21+ 22-K) = K/2*(43-K)
So average of k consecutive numbers = (43 -K )/2

E is the answer

sujoykrdatta · Answer

Largest number = 21
Since there are k consecutive integers, the starting number = 21 - k + 1 = 22 - k
Thus, the average of all the numbers = (1st term + Last term)/2 = (21 + 22 - k)/2 = (43 - k)/2
Answer E

Note: in an arithmetic sequence, mean of all terms = (1st term + last term)/2

shubhim20 · Answer

why are we doing 21+K-1

Bunuel · Answer

In any k consecutive integers, the gap between the smallest and the largest is always k - 1 steps.

So largest = smallest + (k - 1).
 
Thus smallest = largest - (k - 1).

For example, if we have 4 consecutive integers, so if k = 4 and largest = 21, then smallest = 21 - (4 - 1) = 18, and the set is 18, 19, 20, 21.

arushi118 · Answer

Take an example:
If you have three consecutive numbers:
18, 19, 20 -> Now you can write 18 as 20 - 2 -> which is same as 20 - (k-1) where k is the number of consecutive integers that we have.

Similarly, for this question, 
The first term of the list of consecutive numbers can be written as 21 - (k-1) = 21 - k +1 = 22 - k

Note: I think you have written 21 - k + 1 as 21 + k -1 by mistake.

Around the World in 80 Questions (Day 10): In a set of k consecutive

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