39n(n+1)/2=nk
this implies n=2.na-1 where a=k-1
so n should be odd and there exist a suitable `a` satisfying the above equation i gave
Under what condition on n, will n consecutive integers exist whose sum is nk
Hint: Consecutive integers are in AP [6]
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6 Answers
39
6@b555 hw cud u take sum of n integers as n(n+1)/2 .??it isnt said dat it is sum of natural no.........it can start from ny number.......
6let integers be n,n+1,n+2.... till x terms
therefore sum =
{x[2n+(x-1)]}/2=nk.can we proceed like dis?
1I have to say I am not sure what I have done..
sum of n consecutive integers = nA
A being the average of the numbers.
nA = nk
A = nk-1
right side is integer so left side also integer
Now n should be odd because if it's even the average of n consecutive numbers is never a whole number.
I guess that is the only condition on n that it should be odd
and the series is ...nk-1-2,nk-1-1,nk-1,nk-1+1,nk-1+2...
n such numbers with nk-1 in the middle.
62exactly what i had in mind.. (actually i had posted this question long long back..)
sum of an AP.. is the first + last term divided by 2 times the number of terms :)