Chapter 14.5, Problem 80SR

To determine

To find :First three terms of each arithmetic series.

Answer to Problem 80SR

  $13,18,23$

Explanation of Solution

Given information :

We know that, the given information is $n=19,a_n=103,S_n=1102$

Calculation :

The given equation is $S_n=\frac{n}{2}(a_1+a_n)$

Where, a, is first term of series.

  $\begin{array}{l}1102=\frac{19}{2}(a_1+103)\\ \Rightarrow \frac{1102\times 2}{19}=a_1+103\\ \Rightarrow \frac{2204}{19}=a_1+103\\ \Rightarrow 116=a_1+103\\ \Rightarrow a_1=116-103=13\mathrm{...}(1)\end{array}$

And

  $\begin{array}{l}{a}_n=a_1+(n-1)d\\ \Rightarrow 103=13+(19-1)d,\text{Using (1)}\\ \Rightarrow 103-13=(19-1)d\\ \Rightarrow 90=18d\\ \Rightarrow d=5\end{array}$

First three terms, $\text{13, 18, 23}$

  $\begin{array}{l}\text{The given equation is}{\text{S}}_{\text{n}}=\frac{n}{2}(a_1+a_n)\\ \text{where, a, is first term of series}\text{.}\\ 1102=\frac{19}{2}(a_1+103)\\ \Rightarrow \frac{1102\times 2}{19}=a_1+103\\ \Rightarrow \frac{2204}{19}=a_1+103\\ \Rightarrow 116=a_1+103\\ \Rightarrow a_1=116-103=13\mathrm{...}(1)\\ And\\ a_n=a_1+(n-1)d\\ \Rightarrow 103=13+(19-1)d,\text{Using (1)}\\ \Rightarrow 103-13=(19-1)d\\ \Rightarrow 90=18d\\ \Rightarrow d=5\\ \text{First three terms, 13, 18, 23}\end{array}$