Chapter 3, Problem 19PT

To determine

To find: $n^{th}$ term in the sequence.

Answer to Problem 19PT

Given sequence is Arithmetic.

  $n^{th}$ term of the sequence is $a_n=8n-48$ .

Explanation of Solution

Given information:

  $-40,-32,-24,-16,\mathrm{....}$

Concept:

An Arithmetic sequence adds or subtracts a fixed amount (common difference) to get the next term in the sequence.

We know $n^{th}$ term of Arithmetic progression is

  $a_n=a+\left(n-1\right)d$

Where

  $a$ = First term

  $d$ = Common difference

Calculation:

Given series is

  $-40,-32,-24,-16,\mathrm{....}$

Common difference = $d=-32-\left(-40\right)=-24-\left(-32\right)=-16-\left(-24\right)=8$

As difference between any two consecutive terms is same. So, the given series is Arithmetic.

From given sequence

First term = $a=-40$

Common difference = $d=-32-\left(-40\right)=-32+40=8$

Substituting values in $n^{th}$ term formula

We get

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ \Rightarrow a_n=-40+\left(n-1\right)8\\ \Rightarrow a_n=-40+8n-8\\ \Rightarrow a_n=8n-48\end{array}$

Therefore,

  $n^{th}$ term of the sequence is $a_n=8n-48$