Question on: JAMB Mathematics - 1991
What is the nth term of the progression 27, 9, 3,......?
A
27\(\frac{1}{3}\) n - 1
B
3n + 2
C
27 + 18(n - 1)
D
27 + 6(n - 1)
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Correct Option: A
Given 27, 9, 3,......this is a G.P
r = (\frac{9}{27})
= (\frac{1}{3})
T = arn - 1
= 27(\frac{1}{3}) n - 1
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