Question

In Problems 7–12, determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference and the sum of the first n terms. If the sequence is geometric, find the common ratio and the sum of the first n terms.

$3,\frac{3}{2},\frac{3}{4},\frac{3}{8},\frac{3}{16},..$

Short answer

• Given sequence is a geometric sequence.
• Its common ratio is $r=\frac{1}{2}$.
• Sum of first $n$term is $S_n=6\left(1-2^{-n}\right)$.

Step-by-step solution

Step 1. Given information. 

Sequence is:$3,\frac{3}{2},\frac{3}{4},\frac{3}{8},\frac{3}{16},.......$

Step 2. To find common ratio.

The ratio between two successive term is :

$$\begin{gathered} r=\frac{{\displaystyle \frac{3}{2}}}{3} \\ =\frac{1}{2} \end{gathered}$$

$$\begin{gathered} r=\frac{{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{3}{2}}} \\ =\frac{1}{2} \end{gathered}$$

Since all the ratio of successive term in the sequence is constant.

Hence it is a geometric sequence, whose common ratio is$r=\frac{1}{2}$

Step 3. To find sum of first nterms.

First term is $a=3$

common ratio is$r=\frac{1}{2}$

So sum of $n$ terms is:

$$\begin{gathered} S_n=\frac{a(1-r^n)}{1-r} \\ =3.\frac{1-{\left({\displaystyle \frac{1}{2}}\right)}^n}{1-{\displaystyle \frac{1}{2}}} \\ =\frac{3(1-2^{-n})}{{\displaystyle \frac{1}{2}}} \\ =6(1-2^{-n}) \end{gathered}$$