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n consecutive positive integers

Expert replies

by adilka » Sat Dec 13, 2008 7:00 pm
earth@work wrote:
adilka wrote:
earth@work wrote:sum = (a+1) + (a+2) +.....(a+n) for any positive integer a
= na+n(n+1)/2 = 45
now to find the value of n we need to know 'a' which is given in neither of statements...so both insuff
ans E .... do let me know if u see some error here, as this is the best i cud think of without plugging nos.
There is a slight error in your calculations. It should be na+n(n-1)/2=45
i.e. (n-1) not (n+1) the reason for that is that the first number in sequence, which you assumed to be "a" has an adder of 0, hence you have (n-1) numbers that start with 1.
Hi adilka,
my first number of the sequence is (a+1) and not 'a', that is the reason why my sum is na+n(n+1)/2
Makes sense.
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