Chapter 3, Problem 52SGR

To determine

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

Answer to Problem 52SGR

Next three terms of the sequence are $39,36,33$ .

Explanation of Solution

Given information:

  $51,48,45,42,\mathrm{....}$

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

  $51,48,45,42,\mathrm{....}$

Common difference = $d=48-51=45-48=42-45=-3$

From given sequence

First term = $a=51$

Common difference = $d=-3$

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

We get

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ \Rightarrow a_n=51+\left(n-1\right)\left(-3\right)\\ \Rightarrow a_n=51-3n-\left(-3\right)\\ \Rightarrow a_n=51-3n+3\\ \Rightarrow a_n=54-3n\end{array}$

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 $a_n=54-3n$

Substitute $n=5$

We get

  $\begin{array}{l}{a}_n=54-3n\\ \Rightarrow a_5=54-3\left(5\right)\\ \Rightarrow a_5=54-15\\ \Rightarrow a_5=39\end{array}$

Substitute $n=6$

  $\begin{array}{l}{a}_n=54-3n\\ \Rightarrow a_6=54-3\left(6\right)\\ \Rightarrow a_6=54-18\\ \Rightarrow a_6=36\end{array}$

Substitute $n=7$

  $\begin{array}{l}{a}_n=54-3n\\ \Rightarrow a_7=54-3\left(7\right)\\ \Rightarrow a_7=54-21\\ \Rightarrow a_7=33\end{array}$

Therefore,

Next three terms of the sequence are $39,36,33$