Chapter 11.2, Problem 5BCYP

To determine

Find the first three terms of the arithmetic series. $a_1=-24,a_n=288,{\text{S}}_n=5280$

Answer to Problem 5BCYP

The first three terms are: $-24,-16,-8$

Explanation of Solution

Given:

  $a_1=-24,a_n=288,{\text{S}}_n=5280$

Concept Used:

The general equation of nth term is: $a_n=a_1+(n-1)·d$

The general equation of sum of n terms is: ${\text{S}}_{\text{n}}=\frac{n}{2}[2a_1+(n-1)·d]$ or ${\text{S}}_{\text{n}}=\frac{n}{2}(a_1+a_n)$

Calculation:

Here given:$a_1=-24,a_n=288,{\text{S}}_n=5280$

  $\begin{array}{l}{a}_1=-24,a_n=288,{\text{S}}_n=5280\\ {\text{S}}_{\text{n}}=\frac{n}{2}(a_1+a_n)\\ 5280=\frac{n}{2}(-24+288)\\ 5280=\frac{n}{2}·264\\ 5280=132n\\ \frac{5280}{132}=\frac{132n}{132}\\ n=40\end{array}$

  $\begin{array}{l}{a}_1=-24;n=40;a_n=288\\ a_n=a_1+(n-1)·d\\ 288=-24+(40-1)·d\\ 288+24=39d\\ 312=39d\\ \frac{39d}{39}=\frac{312}{39}\\ d=8\end{array}$

  $\begin{array}{l}{a}_1=-24,d=8\\ a_2=a_1+d=-24+8=-16\\ a_3=a_1+2d=-24+16=-8\end{array}$

Thus, the first three terms are: $-24,-16,-8$