Chapter 12.2, Problem 50E

To determine

Calculate the sum of the given partial sum of an arithmetic sequence $-3+\left(-\frac{3}{2}\right)+0+\frac{3}{2}+3+\mathrm{.....}+30$ .

Answer to Problem 50E

The sum of the given partial sum of an arithmetic sequence $-3+\left(-\frac{3}{2}\right)+0+\frac{3}{2}+3+\mathrm{.....}+30$ is $S_{23}=310.5$ .

Explanation of Solution

Given information:

It is given in the question that, the partial sum of an arithmetic sequence is $-3+\left(-\frac{3}{2}\right)+0+\frac{3}{2}+3+\mathrm{.....}+30$ .

Formula used:

  $S_n=n\left[\frac{a+a_n}{2}\right]\&a_n=a+\left(n-1\right)d$

Where,

  $\begin{array}{l}aisthe{1}^{st}term\\ nisthenumberofterm\\ disthecommondifference\end{array}$

Calculation:

The partial sum of an arithmetic sequence is $-3+\left(-\frac{3}{2}\right)+0+\frac{3}{2}+3+\mathrm{.....}+30$ means,

  $\begin{array}{l}{a}_1=-3\Rightarrow a=-3\\ a_2=\left(-\frac{3}{2}\right)\Rightarrow a+d=-\frac{3}{2}\Rightarrow d=-\frac{3}{2}-(-3)\Rightarrow d=\frac{-3+6}{2}\Rightarrow d=\frac{3}{2}\end{array}$

So, $a=-3$ & $d=\frac{3}{2}$

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ 30=-3+\left(n-1\right)\frac{3}{2}\\ 30+3=\left(n-1\right)\frac{3}{2}\\ 33*\frac{2}{3}=\left(n-1\right)\\ 11*2=\left(n-1\right)\\ 22=\left(n-1\right)\\ n=23\end{array}$

Now by putting $a=-3,d=\frac{3}{2}\&n=23$ in the formula $S_n=n\left[\frac{a+a_n}{2}\right]$ ,

  $\begin{array}{l}{S}_n=n\left[\frac{a+a_n}{2}\right]\\ S_{23}=23\left[\frac{-3+30}{2}\right]\\ S_{23}=23\left[\frac{27}{2}\right]\\ S_{23}=11.5*27\\ S_{23}=310.5\end{array}$