JAMB - Mathematics (1991 - No. 23)
What is the nth term of the progression 27, 9, 3,......?
27\(\frac{1}{3}\) n - 1
3n + 2
27 + 18(n - 1)
27 + 6(n - 1)
Explanation
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
r = \(\frac{9}{27}\)
= \(\frac{1}{3}\)
T = arn - 1
= 27\(\frac{1}{3}\) n - 1
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