The sum of the series up to the 12 terms will be 40.5.
What is the sum of a geometric sequence?
Let a₁ be the first term, n be the total number term, and r be a common ratio.
Then the sum of the geometric sequence will be
Sₙ = [a₁ (1 - rⁿ)] / (1 - r)
The series is given below.
27 + 9 + 3 + 1 + ... + 1/6561
The first term is 27 and the common ratio is 1/3.
The number of the term will be
1/6561 = 27 · (1/3)ⁿ⁻¹
1/177147 = (1/3)ⁿ⁻¹
(1/3)¹¹ = (1/3)ⁿ⁻¹
11 = n - 1
n = 12
Then the sum of the series will be
[tex]\rm S_{12} = \dfrac{27 \cdot \left [1 - \left (\frac{1}{3} \right )^{12} \right ]}{ \left ( 1- \frac{1}{3} \right )}[/tex]
Simplify the equation, then we have
S₁₂ = 27 x 0.9999 / 0.6666
S₁₂ = 27 x 1.5
S₁₂ = 40.5
The sum of the series up to the 12 terms will be 40.5.
More about the sum of arithmetic sequence link is given below.
https://brainly.com/question/25749583
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