Chapter 12, Problem 21RE

To determine

To find: Whether the given sequence is arithmetic, geometric or neither. Determine the common difference and the sum of first n terms if the sequence is arithmetic. Determine the common ratio and the sum of First n terms if the sequence is geometric.

Answer to Problem 21RE

The sequence is geometric with common ration of $\frac{1}{2}$ and the sum of first n term is $6\left[1-{\left(\frac{1}{2}\right)}^n\right]$ .

Explanation of Solution

Given:

The given series is $3,\frac{3}{2},\frac{3}{4},\frac{3}{8},\frac{3}{16},\mathrm{....}$ .

Calculation:

Consider the given series is,

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

The difference of the first and the second term is,

  $\frac{3}{2}-3=\frac{-3}{2}$

The difference of the third and the second term is,

  $\frac{3}{4}-\frac{3}{2}=\frac{-3}{4}$

Thus, the sequence is not arithmetic.

The common ratio of the first and the second term is,

  $\frac{\frac{3}{2}}{3}=\frac{1}{2}$

The common ratio of the second and the third term is,

  $\frac{\frac{3}{4}}{\frac{3}{2}}=\frac{1}{2}$

Thus, the sequence is geometric.

The sum of the first n term of the geometric sequence is,

  $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$

Then,

  $\begin{array}{c}{S}_n=3\left(\frac{1-{\left(\frac{1}{2}\right)}^n}{1-\frac{1}{2}}\right)\\ =6\left[1-{\left(\frac{1}{2}\right)}^n\right]\end{array}$