Chapter 12.2, Problem 40E

To determine

To calculate:

Sum of the first ten terms of the given geometric series.

Answer to Problem 40E

Sum of the first eight terms of the given geometric series is $\frac{-160}{1-\sqrt{3}}$ .

Explanation of Solution

Given information:

  $2+2\sqrt{3}+6+\mathrm{...}$

Calculation:

Here given geometric sequence is $2+2\sqrt{3}+6+\mathrm{...}$

  $a_1=2,a_2=2\sqrt{3},a_3=6$

Common ratio is calculated as,

  $r=\frac{a_2}{a_1}=\frac{2\sqrt{3}}{2}=\sqrt{3}$

Where,

  $r=$ Common ratio,

  $a_1=$ First term,

  $a_2=$ Second term.

Sum of first five terms of the given geometric series is calculated as,

  $S_n=a_1.\frac{1-r^n}{1-r}$

Where,

  $S_n=$ Sum of $n^{th}$ term,

  $a_1=$ First term,

  $r=$ Common ratio,

  $n=$ Number of terms,

Put the values in the equation,

  $\begin{array}{l}{S}_8=\frac{2-2{\left(\sqrt{3}\right)}^8}{1-\sqrt{3}}\\ S_8=\left(-92.37\right)\\ S_8=\frac{-160}{1-\sqrt{3}}\end{array}$

Thus, Sum of the first eight terms of the given geometric series is $\frac{-160}{1-\sqrt{3}}$ .