Chapter 12, Problem 11CT

To determine

To find: The common ratio and the sum of the first $n$ terms if the sequence is $\left\{\frac{2n-3}{2n+1}\right\}$ .

Answer to Problem 11CT

The ratio of successive terms is not constant.

The series is neither arithmetic nor geometric.

Explanation of Solution

Given information:

The given sequence is $\left\{\frac{2n-3}{2n+1}\right\}$ .

Calculation:

Substitute $1,2,3,4$ etc for $n$ in $\left\{\frac{2n-3}{2n+1}\right\}$ .

Therefore, $a_1=-\frac{1}{3},a_2=\frac{1}{5},a_3=\frac{3}{7}\text{ and }{a}_4=\frac{5}{9}$ and the sequence is $-\frac{1}{3},\frac{1}{5},\frac{3}{7},\frac{5}{9},\mathrm{....}$

The difference between two successive terms is

  $\begin{array}{c}{a}_2-a_1=\frac{1}{5}-\left(-\frac{1}{3}\right)=\frac{8}{15}\\ a_3-a_2=\frac{3}{7}-\frac{1}{5}=\frac{8}{35}\end{array}$

The difference of successive term is not a constant.

The ratio of successive terms is

  $\begin{array}{c}\frac{a_2}{a_1}=\frac{1}{5}÷\left(-\frac{1}{3}\right)=-\frac{3}{5}\\ \frac{a_3}{a_2}=\frac{3}{7}÷\frac{1}{5}=\frac{15}{7}\end{array}$

The ratio of successive terms is not constant.

Therefore, the series is neither arithmetic nor geometric.