Chapter 12, Problem 9CT

To determine

To find: The common ratio and the sum of the first $n$ terms.

Answer to Problem 9CT

The sum of $n$ terms of the sequence is $s_n=\frac{2}{3}\left(1-4^n\right)$ .

Explanation of Solution

Given information:

The sum of the first $n$ terms of arithmetic sequence is $s_n=\frac{n}{4}\left(27-n\right)$ .

Calculation:

Substitute the values of $1,2,3$ and $4$ for $n$ in the series $\left\{-\frac{n}{2}+7\right\}$ .

Thus, $a_1=\frac{13}{2}$ , $a_2=6$ , $a_3=\frac{11}{2}$ and $a_4=5$ .

The sequence is $\frac{13}{2},6,\frac{11}{2}\mathrm{.5...............}$

Calculate the difference between successive terms.

  $a_2-a=6-\frac{13}{2}=-\frac{1}{2}$

  $a_3-a_2=\frac{11}{2}-6=-\frac{1}{2}$

The difference between successive terms is not constant and is $-\frac{1}{2}$

The ratio of the successive terms is

  $\frac{a_2}{a_1}=6÷\frac{13}{2}=\frac{12}{13}$

  $\frac{a_1}{a_2}=\frac{11}{2}÷6=\frac{11}{12}$

Thus, the ratio of successive terms is not constant.

Therefore, the series is arithmetic and the common difference is $-\frac{1}{2}$

Substitute the values of $\frac{13}{2}$ for $a_1$ and $-\frac{1}{2}$ for $d$ in formula,

  $\begin{array}{c}{S}_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]\\ =\frac{n}{2}\left[2\times \frac{13}{2}+\left(n-1\right)\left(-\frac{1}{2}\right)\right]\\ =\frac{n}{2}\left(13-\frac{n}{2}+\frac{1}{2}\right)\\ =\frac{n}{4}\left(27-n\right)\end{array}$

Therefore, the sum of the first $n$ terms of arithmetic sequence is $s_n=\frac{n}{4}\left(27-n\right)$ .