Chapter 7.2, Problem 45E

Solution

To calculate: The sum of the arithmetic series ${\displaystyle \sum _{i=5}^{14}\left(-54+9i\right)}$

  ${\displaystyle \sum _{i=5}^{14}\left(-54+9i\right)}=315$

Given information:

The given terms are ${\displaystyle \sum _{i=5}^{14}\left(-54+9i\right)}$ .

Formula used:

The sum nth term of an arithmetic sequence with first term $a_1$ and $a_4$ is given by: $s_n=n\left(\frac{a_1+a_n}{2}\right)$

Calculation:

Consider the given sequence.

It is given that ${\displaystyle \sum _{i=5}^{14}\left(-54+9i\right)}$

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

The first term

  $\begin{array}{l}{a}_1=-54+9\times 1\\ a_1=-54+9\\ a_1=-45\end{array}$

The fourth term

  $\begin{array}{c}{a}_4=-54+9\times 4\\ a_4=-54+36\\ a_4=-18\end{array}$

The last term

  $\begin{array}{c}{a}_{14}=-54+9\times 14\\ a_{14}=-54+126\\ a_{14}=72\end{array}$

The given sequence is arithmetic with first term $a_1=-45\text{and}{a}_4=-18$

  $\begin{array}{c}{s}_4=n\left(\frac{a_1+a_4}{2}\right)\\ =\frac{4\left(-45+\left(-18\right)\right)}{2}\\ =-126\end{array}$

Sum of first 14 terms (from $i=1\text{to}14$ )

The given sequence is arithmetic with first term $a_1=-45\text{and}{a}_{14}=72$

  $\begin{array}{c}{s}_{14}=n\left(\frac{a_1+a_{14}}{2}\right)\\ =\frac{4\left(-45+\left(72\right)\right)}{2}\\ =189\end{array}$

Now the subtract to get the sum from $5\text{to}14$

  $189-\left(-126\right)=315$