Chapter 3, Problem 50SGR

To determine

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

Answer to Problem 50SGR

Next three terms of the sequence are $31,38,45$ .

Explanation of Solution

Given information:

  $3,10,17,24,\mathrm{......}$

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

  $3,10,17,24,\mathrm{......}$

Common difference = $d=10-3=17-10=24-17=7$

From given sequence

First term = $a=3$

Common difference = $d=7$

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

We get

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ \Rightarrow a_n=3+\left(n-1\right)\left(7\right)\\ \Rightarrow a_n=3+7n-7\\ \Rightarrow a_n=7n+3-7\\ \Rightarrow a_n=7n-4\end{array}$

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 $a_n=7n-4$

Substitute $n=5$

We get

  $\begin{array}{l}{a}_n=7n-4\\ \Rightarrow a_5=7\left(5\right)-4\\ \Rightarrow a_5=35-4\\ \Rightarrow a_5=31\end{array}$

Substitute $n=6$

  $\begin{array}{l}{a}_n=7n-4\\ \Rightarrow a_6=7\left(6\right)-4\\ \Rightarrow a_6=42-4\\ \Rightarrow a_6=38\end{array}$

Substitute $n=7$

  $\begin{array}{l}{a}_n=7n-4\\ \Rightarrow a_7=7\left(7\right)-4\\ \Rightarrow a_7=49-4\\ \Rightarrow a_7=45\end{array}$

Therefore,

Next three terms of the sequence are $31,38,45$