Chapter 12.3, Problem 72AYU

To determine

To find: Whether the sequence is arithmetic, geometric or neither, find the common difference or common ratio and also find the sum of the first fifty terms if the sequence is arithmetic or geometric.

Answer to Problem 72AYU

The sequence is arithmetic, the common difference is $\frac{-3}{4}$ and the sum of first $50$ terms is $S_{50}=-556.25$ .

Explanation of Solution

Given information:

The series is $\left\{8-\frac{3}{4}n\right\}$ .

Calculation:

Assume $a_n=\left\{8-\frac{3}{4}n\right\}$ .

Calculate the first term.

  $\begin{array}{l}{a}_1=8-\frac{3}{4}\left(1\right)\\ a_1=\frac{29}{4}\end{array}$

Calculate nth term.

  $a_n=\left\{8-\frac{3}{4}n\right\}$

Calculate $\left(n-1\right)\text{th}$ term.

  $a_{n-1}=8-\frac{3\left(n-1\right)}{4}$

Calculate the difference of nth term and $\left(n-1\right)\text{th}$ term.

  $\begin{array}{c}d=a_n-a_{n-1}\\ =8-\frac{3}{4}n-\left(8-\frac{3\left(n-1\right)}{4}\right)\\ =\frac{-3}{4}\end{array}$

Since the difference of consecutive terms is constant, the sequence is arithmetic.

Calculate the sum of first $50$ terms.

  $\begin{array}{l}{S}_n=\frac{n}{2}\left(a+l\right)\\ S_{50}=\frac{50}{2}\left(\frac{29}{4}+\left(8-\frac{3}{4}n\right)\right)\\ S_{50}=25\left(\frac{29}{4}+\left(8-\frac{3}{4}\left(50\right)\right)\right)\\ S_{50}=-556.25\end{array}$

Therefore, the sequence is arithmetic, the common difference is $\frac{-3}{4}$ and the sum of first $50$ terms is $S_{50}=-556.25$ .