gracie
The square of the median of a sequence of consecutive positive odd integers is equal to the difference between the squares of the first and last terms. If the sequence has twelve terms, what is the sum of the digits of the first and last terms?
A. 12
B. 16
C. 20
D. 24
E. 28
source:self
The median of 12 consecutive odd integers is between the 6th and 7th terms, where if \(n > 1\), then
\(A_{n} = A_1 + 2(n-1)\)
\(A_1 = n\)
\(A_2 = n + 2\)
.
.
\(A_6 = n + 10\)
\(A_7 = n + 12\)
.
.
\(A_{12} = n + 22\)
Median = \(n + 11\). If not sure:
\(\frac{(n+10+n+12)}{2}=\frac{(2n+22)}{2}=(n + 11)\)
Square of median = difference between squares of first and last terms:
\((n + 22)^2 - (n)^2 = (n + 11)^2\)
\(n^2 + 44n + 484 - n^2 = n^2 + 22n + 121\)
\(n^2 - 22n - 363 = 0\)
\((n - 33)(n + 11) = 0\)
\(n = 33\) (prompt says positive)
\(A_1 = 33\)
\(A_{12} = n + 22 = 55\)
Sum of digits of \(33\) and \(55\):
\(3 + 3 + 5 + 5 = 16\)
ANSWER B