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Rangpur Secondary 2011/1

Posted: Wed Feb 02, 2011 6:45 pm
by Moon
Problem 1:
The sum of $81$ consecutive integers is $9^5$. What is their median?

Re: Rangpur Secondary 2011/1

Posted: Wed Feb 02, 2011 6:50 pm
by Hasib
$(x-40)+(x-39)....+x+....(x+40)=9^5$
so, $81x=9^5$
so, $x=9^3$

median is $9^3$

Re: Rangpur Secondary 2011/1

Posted: Wed Feb 02, 2011 7:19 pm
by Tahmid Hasan
i did the same way as hasib :D

Re: Rangpur Secondary 2011/1

Posted: Thu Feb 03, 2011 8:20 pm
by *Mahi*
Yup....it is $9^3=729$

Re: Rangpur Secondary 2011/1

Posted: Fri Jan 06, 2012 6:06 pm
by nafistiham
It is true that, when $n=2k + 1$ the summation of any $n$ consecutive positive numbers is equal to $mn$ where $m$ is the median. :D