The sum of the series up to the 12 terms will be 40.5.
<h3>What is the sum of a geometric sequence?</h3>
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
![\rm S_{12} = \dfrac{27 \cdot \left [1 - \left (\frac{1}{3} \right )^{12} \right ]}{ \left ( 1- \frac{1}{3} \right )}](https://tex.z-dn.net/?f=%5Crm%20S_%7B12%7D%20%3D%20%5Cdfrac%7B27%20%5Ccdot%20%20%5Cleft%20%5B1%20-%20%20%5Cleft%20%28%5Cfrac%7B1%7D%7B3%7D%20%5Cright%20%29%5E%7B12%7D%20%20%20%5Cright%20%5D%7D%7B%20%5Cleft%20%28%201-%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright%20%29%7D)
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.
brainly.com/question/25749583
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