Chapter 7.8, Problem 1P

To determine

Todetermine:The geometrical sequence out of four given series. If sequence is not geometric is it arithmetic.

Answer to Problem 1P

The sequences aandcare geometric sequences, sequence bis an arithmetic sequence and sequence d neither a geometric sequence nor an arithmetic sequence.

Explanation of Solution

Given information:

The given series are

a. $3,6,12,24,48,\dots$

b. $3,6,9,12,15,\dots$

c. $\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81},\dots$

d. $4,7,11,16,22,\dots$

Conceptused:

Arithmetic sequence: $a,a+d,a+2d,a+3d,\dots$ , where a is first term and dthe common difference.

Geometric sequence: $a,ar,a{r}^2,a{r}^3,a{r}^4,\dots$ , where a is the first term and r the common ratio.

Calculations:

(a) The sequence $3,6,12,24,48,\dots$ may be written as $3,3\cdot 2,3\cdot 2^2,3\cdot 2^3,{3.2}^4,\dots$ .

Thus, the sequence $3,6,12,24,48,\dots$ is a geometric sequence.

(b). Thesequence $3,6,9,12,15,\dots$ may be written as $3,\left(3+3\right),\left(3+2\cdot 3\right),\left(3+3\cdot 3\right),\left(3+4\cdot 3\right),\dots$

Thus, the sequence $3,6,9,12,15,\dots$ is anarithmetic sequence.

(c). Thesequence $\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81},\dots$ may be written as $\frac{1}{3},\frac{1}{3}\cdot {\left(\frac{1}{3}\right)}^1,\frac{1}{3}\cdot {\left(\frac{1}{3}\right)}^2,\frac{1}{3}\cdot {\left(\frac{1}{3}\right)}^3,\dots$

Thus, the sequence $3,6,9,12,15,\dots$ is a geometric sequence.

(d). Thesequence $4,7,11,16,22,\dots$ may be written as $4,4+3,4+7,4+12,4+15,\dots$

Thus, the sequence $4,7,11,16,22,\dots$ is neither a geometric sequence nor an arithmetic sequence.

Conclusion:

The sequences aandcare geometric sequences, sequence bis an arithmetic sequence and sequence d neither a geometric sequence nor an arithmetic sequence.