Chapter 3, Problem 48SGR

To determine

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

Answer to Problem 48SGR

Next three terms of the sequence are $0.8,0.6,0.4$ .

Explanation of Solution

Given information:

  $1.4,1.2,1.0,\mathrm{.....}$

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

  $1.4,1.2,1.0,\mathrm{.....}$

Common difference = $d=1.2-1.4=1.0-1.2=-0.2$

From given sequence

First term = $a=1.4$

Common difference = $d=-0.2$

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

We get

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ \Rightarrow a_n=1.4+\left(n-1\right)\left(-0.2\right)\\ \Rightarrow a_n=1.4-0.2n-\left(-0.2\right)\\ \Rightarrow a_n=1.4-0.2n+0.2\\ \Rightarrow a_n=1.6-0.2n\end{array}$

As given first three terms of the series.

To find next three terms we need to substitute $n=4,5,6$

Therefore,

  $n^{th}$ term of the sequence is $a_n=1.6-0.2n$

Substitute $n=4$

We get

  $\begin{array}{l}{a}_n=1.6-0.2n\\ \Rightarrow a_4=1.6-0.2\left(4\right)\\ \Rightarrow a_4=1.6-0.8\\ \Rightarrow a_4=0.8\end{array}$

Substitute $n=5$

  $\begin{array}{l}{a}_n=1.6-0.2n\\ \Rightarrow a_5=1.6-0.2\left(5\right)\\ \Rightarrow a_5=1.6-1.0\\ \Rightarrow a_5=0.6\end{array}$

Substitute $n=6$

  $\begin{array}{l}{a}_6=1.6-0.2n\\ \Rightarrow a_6=1.6-0.2\left(6\right)\\ \Rightarrow a_6=1.6-1.2\\ \Rightarrow a_6=0.4\end{array}$

Therefore,

Next three terms of the sequence are $0.8,0.6,0.4$