Question

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

Answer 1

517

Explanation:

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

The sum of `k` consecutive odd integers, starting at the odd number with index `n`, 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.