GeminiHeat wrote:
In a certain sequence, the term \(A_n\) is given by the formula \(A_n= 16-\frac{(4- n)^3}{|n-4|}\) for all positive integers n ≠ 4, while \(A_4 = 16\). For which integer value of k greater than 2 is the mean of all values in the sequence from \(A_1\) through \(A_k\) equal to the median of those values?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
\(A_n= 16-\frac{(4-n)^3}{|n-4|}\)
\(A_1 = 7\)
\(A_2 = 12\)
\(A_3 = 15\)
\(A_4 = 16\)
\(A_5 = 17\)
\(A_6 = 20\)
\(A_7 = 25\)
\(A_8 = 32\)
\(A_9 = 41\)
I. \(k = 6\)\(\frac{7+12+15+16+17+20}{6} = \frac{15+16}{2}\)
\(14.5 ≠ 15.5\)
II. \(k = 7\)\(\frac{7+12+15+16+17+20+25}{7} = 16\)
\(16 = 16\)
Hence, option B