Q1, Q2, Q3, …, Qn is a sequence of n consecutive multiples of positive integer k. If n > 4, then what is the average (arithmetic mean) of the sequence?
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\) = ?
(1) k = 4
Case1: Let \(Q_1 = k, Q_2 = 2k, Q_3 = 3k ... Q_{10} = 10k\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{k + 2k + 3k + 4k + .... 10k}{10}\)
\(Avg. = \frac{11}{2}*k = 22\)
Case 2: Let \(Q_1 = 5k, Q_2 = 6k, Q_3 = 7k ... Q_5 = 9k\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{5k + 6k + 7k + 8k + 9k}{5}\)
Avg. = 7k = 28
INSUFFICIENT.
(2) The sum of the second greatest and the second least terms of the series is 24.
\(Q_2 + Q_{n-1} = 24\)
Case 1: Let \(Q_1 = k, Q_2 = 2k, Q_3 = 3k ... + Q_{10} = 10k\)
2k + 9k = 24
\(k = \frac{24}{11}\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 ..... Q_{n-1} + Q_n}{n}\)
\(Avg. = \frac{k + 2k + 2k + 4k + ..... + 10k}{10}\)
\(Avg. = \frac{11}{2}k = \frac{11}{2}*\frac{24}{11} = 12\)
Case 2: Let \(Q_1 = 5k, Q_2 = 6k, Q_3 = 7k ... Q_5 = 9k\)
6k + 8k = 24
\(k = \frac{24}{14}\)
\(Avg. = \frac{Q_1 + Q_2 + Q_3 + Q_4 + Q_5}{5}\)
\(Avg. = \frac{5k + 6k + 7k + 8k + 9k}{5}\)
\(Avg. = 7k = 7*\frac{24}{14} = 12 \)
SUFFICIENT.
Answer B.
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