Chapter 9, Problem 32RE

Solution

To state: The first six terms and the 12^th term of the geometric sequence:

  $a_1=3\text{and}r=\frac{1}{3}$

The resultants first six terms are $3,1,\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81}$ and the 12^th term is $\frac{1}{59,049}$ .

Given information:

The given first term is $a_1=3$ and common ratio is $r=\frac{1}{3}$ .

Explanation:

The first term of the sequence is: $a_1=3$ .

Common ratio is: $r=\frac{1}{3}$

The nth term of the geometric series is:

  $a_n=a_1\cdot r^{n-1}$

Now substitute 2 for n for the second term and simplify,

  $\begin{array}{c}{a}_2=a_1\cdot {\left(\frac{1}{3}\right)}^{2-1}\\ =3\left(\frac{1}{3}\right)\\ =1\end{array}$

Now substitute 3 for n for the third term and simplify,

  $\begin{array}{c}{a}_3=a_1\cdot {\left(\frac{1}{3}\right)}^{3-1}\\ =3{\left(\frac{1}{3}\right)}^2\\ =\frac{1}{3}\end{array}$

Similarly, find the remaining 3 terms,

  $\begin{array}{c}{a}_4=a_1\cdot {\left(\frac{1}{3}\right)}^{4-1}\\ =3{\left(\frac{1}{3}\right)}^3\\ =\frac{1}{9}\\ a_5=a_1\cdot {\left(\frac{1}{3}\right)}^{5-1}\\ =3{\left(\frac{1}{3}\right)}^4\\ =\frac{1}{27}\\ a_6=a_1\cdot {\left(\frac{1}{3}\right)}^{6-1}\\ =3{\left(\frac{1}{3}\right)}^5\\ =\frac{1}{81}\end{array}$

The first six terms are $3,1,\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81}$ .

Then, the 12^th term is:

  $\begin{array}{c}{a}_{12}=3{\left(\frac{1}{3}\right)}^{12-1}\\ =5{\left(\frac{1}{3}\right)}^{11}\\ =\frac{1}{59,049}\end{array}$

Therefore, the first six terms are $3,1,\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81}$ and the 12^th term is $\frac{1}{59,049}$ .