Chapter 6.5, Problem 12E

To determine

To find:Whether the sequence $\frac{3}{49},\frac{3}{7},3,\mathrm{21...}$ is arithmetic, geometric or neither.

Answer to Problem 12E

The sequence $\frac{3}{49},\frac{3}{7},3,\mathrm{21...}$ is geometric.

Explanation of Solution

Given information:The sequence $\frac{3}{49},\frac{3}{7},3,\mathrm{21...}$

Calculation:A sequence is said to be arithmetic sequence if every term of the sequence, after the first term, is obtained by adding a constant to the preceding term. The constant is known as common difference.

A sequence is said to be geometric sequence if every term of the sequence, after the first term, is obtained by multiplying a constant to the preceding term. The constant is known as common ratio.

Here, the 1^st term $=\frac{3}{49}$

Difference between 1^st and 2^nd term $=\frac{3}{7}-\frac{3}{49}=\frac{18}{49}$

Difference between 2^nd and 3^rd term $=3-\frac{3}{7}=\frac{18}{7}$

Difference between 3^rd and 4^th term $=21-3=18$

Since, each term of the sequence is not obtained by adding a constant to the preceding term, the sequence is not arithmetic.

Again, the 1^st term $=\frac{3}{49}$

Dividing 2^nd term by 1^st term $=\frac{3}{7}÷\frac{3}{49}=\frac{3}{7}\times \frac{49}{3}=7$

Dividing 3^rd term by 2^nd term $=3÷\frac{3}{7}=3\times \frac{7}{3}=7$

Dividing 4^th term by 3^rd term $=21÷3=7$

And so on.

Since, each term of the sequence is obtained by multiplying a constant 7 to the preceding term, the sequence is geometric.