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

1 Answer
Apr 2, 2017

Answer:

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.