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

Apr 2, 2017

517

#### Explanation:

Write the $n : t h$ odd number as ${a}_{n} = \left(2 n - 1\right)$, 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
${s}_{n , k} = \left(2 n - 1\right) + \left(2 \left(n + 1\right) - 1\right) + \ldots + \left(2 \left(n - \left(k - 1\right)\right) - 1\right)$
and simplified to
${s}_{n , k} = k \left(2 n - 1\right) + 2 \left(0 + 2 + 4 + \ldots + \left(k - 1\right)\right)$
${s}_{n , k} = k \left(2 n - 1\right) + 2 \left({\sum}_{i = 0}^{k - 1} i\right)$
${s}_{n , k} = k \left(2 n - 1\right) + 2 \left({\sum}_{i = 1}^{k} \left(i - 1\right)\right)$,
by grouping terms and changing summation index. Knowing that the rightmost sum is an arithmetic sum , ${s}_{n , k}$ can now be written
${s}_{n , k} = k \left(2 n - 1\right) + 2 \frac{k \left(k - 1\right)}{2}$
${s}_{n , k} = k \left(2 n - 1\right) + {k}^{2} - k$
${s}_{n , k} = k \left(2 n + k - 2\right)$

We know that the average of the elements in ${s}_{n , k}$ should be $534$, which says that
$\frac{{s}_{n , k}}{k} = 534$
$\frac{k \left(2 n + k - 2\right)}{k} = 534$
$2 n - 1 = 534 - \left(k - 1\right)$.

From this expression we read off the first number in the sequence, for $k = 18$ as
${a}_{n} = 2 n - 1 = 534 - \left(18 - 1\right) = 517$,

which is the sought first number in the sequence.