Question

81, 27, 9, 3......is a geometric progression with common ration=1/3
The 10th term of this geometric progression?

Answer 1

Answer:

$\frac{1}{243}$

Step-by-step explanation:

The nth term, by observation, is:

$t_{n} = \frac{243}{3^n}$

So, putting n = 10, we get:

$t_{10} = \frac{3^5}{3^{10}} = \frac{1}{3^5} = \frac{1}{243}$

Answer 2

Answer: 121.5

Step-by-step explanation:

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The 10th term of this geometric progression= 81×$(1/3)^{10-1}$

=$3^{-5}$

$Thesumtoinfinityofthisgeometricprogression$

=81÷(1-1/3)

=121.5

(^_-) There u go....