Chapter 3, Problem 47SGR

To determine

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

Answer to Problem 47SGR

Next three terms of the sequence are $26,31,36$

Explanation of Solution

Given information:

  $6,11,16,21,\mathrm{.....}$

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

  $6,11,16,21,\mathrm{.....}$

Common difference = $d=11-6=16-11=21-16=5$

From given sequence

First term = $a=6$

Common difference = $d=5$

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

We get

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ \Rightarrow a_n=6+\left(n-1\right)5\\ \Rightarrow a_n=6+5n-5\\ \Rightarrow a_n=5n+1\end{array}$

As given first four 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 $a_n=5n+1$

Substitute $n=5$

We get

  $\begin{array}{l}{a}_n=5n+1\\ \Rightarrow a_5=5\left(5\right)+1\\ \Rightarrow a_5=25+1\\ \Rightarrow a_5=26\end{array}$

Substitute $n=6$

  $\begin{array}{l}{a}_n=5n+1\\ \Rightarrow a_6=5\left(6\right)+1\\ \Rightarrow a_6=30+1\\ \Rightarrow a_6=31\end{array}$

Substitute $n=7$

  $\begin{array}{l}{a}_n=5n+1\\ \Rightarrow a_7=5\left(7\right)+1\\ \Rightarrow a_7=35+1\\ \Rightarrow a_7=36\end{array}$

Therefore,

Next three terms of the sequence are $26,31,36$