The sum of 49 consecutive integers is 7^5. What is their median?

Question

(A) 7
(B) 7^2
(C) 7^3
(D) 7^4
(E) 7^5

KSBGC · Answer

49 is an odd number. thus median = mean.

Sum is given.

Average/median:  \frac{7^5}{49}

\frac{7^5}{7^2}

7^3

C is the correct answer.

gracie · Answer

median of consecutive sequence=sum of sequence/number of terms
7^5/7^2=7^3
C

Archit3110 · Answer

solve the given information
7^5/7^2
IMO C
7^3

PriyankaPalit7 · Answer

For consecutive numbers, median = mean

Mean = Sum of all integers / number of integers

Mean =  7^5 /49 =  7^5 / 7^2  =  7^3

Answer C .

BelalHossain046 · Answer

In case of consecutive numbers(and evenly spaced sets), mean=median.

So here, Median=Mean =(sum/number of terms)=(7^5/49)=(7^5/7^2)
=7^(5-3)=7^2.

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ScottTargetTestPrep · Answer

In a set of evenly spaced integers, the median is equal to the average.

Since average x quantity = sum, we see that average = median = sum/quantity, thus:

Median = 7^5/49 = 7^5/7^2 = 7^3

Answer: C

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