Chapter 3, Problem 51SGR

To determine

To find: Next three terms in the given arithmetic sequence.

Answer to Problem 51SGR

Next three terms of the sequence are $2,6,10$ .

Explanation of Solution

Given information:

  $-14,-10,-6,-2,\mathrm{.....}$

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

  $-14,-10,-6,-2,\mathrm{.....}$

Common difference = $d=-10-\left(-14\right)=-6-\left(-10\right)=-2-\left(-6\right)=4$

From given sequence

First term = $a=-14$

Common difference = $d=4$

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

We get

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ \Rightarrow a_n=-14+\left(n-1\right)\left(4\right)\\ \Rightarrow a_n=-14+4n-4\\ \Rightarrow a_n=4n-14-4\\ \Rightarrow a_n=4n-18\end{array}$

As given first three terms of the series.

To find next three terms we need to substitute $n=5,6,7$

Therefore,

  $n^{th}$ term of the sequence is $a_n=4n-18$

Substitute $n=5$

We get

  $\begin{array}{l}{a}_n=4n-18\\ \Rightarrow a_5=4\left(5\right)-18\\ \Rightarrow a_5=20-18\\ \Rightarrow a_5=2\end{array}$

Substitute $n=6$

  $\begin{array}{l}{a}_n=4n-18\\ \Rightarrow a_6=4\left(6\right)-18\\ \Rightarrow a_6=24-18\\ \Rightarrow a_6=6\end{array}$

Substitute $n=7$

  $\begin{array}{l}{a}_n=4n-18\\ \Rightarrow a_7=4\left(7\right)-18\\ \Rightarrow a_7=28-18\\ \Rightarrow a_7=10\end{array}$

Therefore,

Next three terms of the sequence are $2,6,10$