2009 AMC 12B Problem 12

Attempt Problem 12 of the 2009 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

The fifth and eighth terms of a geometric sequence of real numbers are 7!7! and 8!8! respectively. What is the first term?
6060
7575
120120
225225
315315
Answer: E
Concepts:geometric sequencefactorial
Small hint:
The ratio of the eighth term to the fifth term is r3r^3
Big hint:
8!7!=8=r3,\dfrac{8!}{7!} = 8 = r^3, so r=2r = 2; then divide 7!7! by r4r^4
Solution:
The eighth term divided by the fifth term is r3=8!7!=8,r^3 = \dfrac{8!}{7!} = 8, so r=2.r = 2.
The fifth term is ar4=7!,a r^4 = 7!, so a=7!16=504016=315.a = \dfrac{7!}{16} = \dfrac{5040}{16} = 315.
Thus, the correct answer is E.
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