Chapter 7.2, Problem 42E

Solution

To calculate: The sum of the arithmetic series ${\displaystyle \sum _{i=1}^{18}\left(14-6i\right)}$

  ${\displaystyle \sum _{i=1}^{18}\left(14-6i\right)}=-774$

Given information:

The given terms are ${\displaystyle \sum _{i=1}^{18}\left(14-6i\right)}$ .

Formula used:

The sum nth term of an arithmetic sequence with first term $a_1$ and $a_{18}$ is given by: $s_n=n\left(\frac{a_1+a_n}{2}\right)$

Calculation:

Consider the given sequence.

It is given that ${\displaystyle \sum _{i=1}^{18}\left(14-6i\right)}$

Put various value for $i$ and solve for first and last term.

The first term

  $\begin{array}{l}{a}_1=14-6\times 1\\ a_1=14-6\\ a_1=8\end{array}$

The last term

  $\begin{array}{l}{a}_{18}=14-6\times 18\\ a_{18}=14-108\\ a_{18}=-94\end{array}$

The given sequence is arithmetic with first term $a_1=8\text{and}{a}_{18}=-94$

  $\begin{array}{c}{s}_{18}=n\left(\frac{a_1+a_{18}}{2}\right)\\ =\frac{18(8-94)}{2}\\ =-774\end{array}$