Chapter 11, Problem 20CLR

To determine

To calculate:

The series is arithmetic or geometric or neither and sum of ${\displaystyle \sum _{n=1}^{11}\left(35-3n\right)}$

Answer to Problem 20CLR

The sequence is arithmetic and sum of ${\displaystyle \sum _{n=1}^{11}\left(35-3n\right)}$ is $187$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{11}\left(35-3n\right)}$

  $n=1,2,3,\cdot \cdot \cdot ,11$

Formula used:

  $a_n=a_1\cdot r^{n-1}$

Calculation:

Consider the series:

  ${\displaystyle \sum _{n=1}^{11}\left(35-3n\right)}\mathrm{...}\left(1\right)$

Put the value of $n=1,2,3,\cdot \cdot \cdot ,11$ in series

  $\begin{array}{l}={\displaystyle \sum _{n=1}^{11}\left(35-3n\right)}\\ =\left(35-3\left(1\right)\right)+\left(35-3\left(2\right)\right)+\left(35-3\left(3\right)\right)+\left(35-3\left(4\right)\right)+\left(35-3\left(5\right)\right)+\left(35-3\left(6\right)\right)+\left(35-3\left(7\right)\right)+\left(35-3\left(8\right)\right)\\ +\left(35-3\left(9\right)\right)+\left(35-3\left(10\right)\right)+\left(35-3\left(11\right)\right)\\ =32+29+26+23+20+17+14+11+8+5+2\end{array}$

The common difference of the sequence:

  $\begin{array}{l}d=a_2-a_1=29-32=-3\\ d=a_3-a_2=26-29=-3\\ d=a_4-a_3=20-23=-3\\ d=a_5-a_4=17-20=-3\\ d=a_6-a_5=14-17=-3\\ d=a_7-a_6=11-14=-3\\ d=a_8-a_7=8-11=-3\\ d=a_9-a_8=5-8=-3\\ d=a_{10}-a_9=2-5=-3\end{array}$

This is the arithmetic sequence. The common difference is $d=-3$

  $\begin{array}{l}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_n=\frac{11}{2}\left(32+2\right)\\ S_n=5.5\left(34\right)\\ S_n=187\end{array}$