If the average of 18 consecutive odd integers is 534, what is the least of these integers?

1 Answer
Apr 2, 2017

517

Explanation:

Write the n:thn:th odd number as a_n = (2n-1)an=(2n1), where nn is an integer.

The sum of kk consecutive odd integers, starting at the odd number with index nn, can then be written as
s_(n,k) = (2n - 1) + (2(n+1) - 1) + ... + (2(n - (k-1)) - 1)
and simplified to
s_(n,k) = k(2n - 1) + 2(0 + 2 +4 + ... + (k-1))
s_(n,k) = k(2n - 1) + 2(sum_(i=0)^(k-1) i)
s_(n,k) = k(2n - 1) + 2(sum_(i=1)^(k) (i-1)),
by grouping terms and changing summation index. Knowing that the rightmost sum is an arithmetic sum , s_(n,k) can now be written
s_(n,k) = k(2n - 1) + 2(k(k-1))/2
s_(n,k) = k(2n - 1) + k^2 - k
s_(n,k) = k(2n +k - 2)

We know that the average of the elements in s_(n,k) should be 534, which says that
(s_(n,k))/k = 534
(k(2n + k - 2))/k = 534
2n - 1 = 534 - (k - 1).

From this expression we read off the first number in the sequence, for k=18 as
a_n = 2n-1 = 534 - (18 - 1) = 517,

which is the sought first number in the sequence.