Question

what is the sum of the first four terms of a geometric series with 2 as its first term and a common ratio of 1/3​

Answer 1

`\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle S_n=\sum\limits_(i=1)^(n)\ a_1\cdot r^(i-1)\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\ \cline{1-1} n=4\\ a_1=2\\ r=(1)/(3) \end{cases}`

`\bf S_4=2\left( \cfrac{1-\left( (1)/(3) \right)^4}{1-(1)/(3)} \right)\implies S_4 = 2\left( \cfrac{1-(1)/(81)}{(2)/(3)} \right)\implies S_4 = 2\left( \cfrac{(80)/(81)}{~~(2)/(3)~~} \right) \\\\\\ S_4=2\left( \cfrac{40}{27} \right)\implies S_4=\cfrac{80}{27}\implies S_4=2(26)/(27)`

Answer 2

80/27.

Explanation:

Sum of n terms = a1 * (1 - r^n) / (1 - r)

Sum of 4 terms = 2 * (1 -(1/3)^4) / ( 1 - 1/3)

= 2 * 80/81 / 2/3

= 160 / 81 * 3/2

= 480/ 162

= 80/27 (answer