Chapter 11.2, Problem 16WE

To determine

To calculate:

The value of $t_{40}$ for following arithmetic sequence:

  $t_8=60,t_{12}=48$

Answer to Problem 16WE

The value of $t_{40}$ the arithmetic sequence is $-36$ .

Explanation of Solution

Given information:

  $t_8=60,t_{12}=48$

Formula used:

The $n^{th}$ term of an arithmetic sequence is given by the following expression:

  $t_n=t_1+\left(n-1\right)d$

Where,

First term: $t_1$

Common difference: $d$

Calculation:

The equations are formed as:

  $\begin{array}{l}{t}_8=t_1+\left(8-1\right)d\\ 60=t_1+7d\end{array}$

  $t_1+7d=60$  ... (1)

And

  $\begin{array}{l}{t}_{12}=t_1+\left(12-1\right)d\\ 48=t_1+11d\end{array}$

  $t_1+11d=48$  ... (2)

Now, $t_1$ is calculated by taking subtraction of both eqns.

  $\begin{array}{l}{t}_1+7d-t_1-11d=60-48\\ -4d=12\\ d=-3\end{array}$

Now, put value of $d=-3$ in eq. (1):

  $\begin{array}{l}{t}_1+7d=60\\ t_1+7\left(-3\right)=60\\ t_1=60+21\\ t_1=81\end{array}$

Now, value of $t_{40}$ the arithmetic sequence is calculated as:

  $\begin{array}{l}{t}_{40}=t_1+\left(40-1\right)d\\ t_{40}=81+39\left(-3\right)\\ t_{40}=39-117\\ t_{40}=-36\end{array}$