Chapter 7, Problem 36PT

To determine

To Find: The first five terms of each sequence $a_1=18,a_n=a_{n-1}-4,n\ge 2$

Answer to Problem 36PT

First five terms of the sequence are $a_1=18,a_2=14,a_3=10,a_4=6\&a_5=2$

Explanation of Solution

Given information:

  $a_1=18,a_n=a_{n-1}-4,n\ge 2$

Concept and Formula Used:

To find the terms put the value of ‘n’.

Calculation:

It is given that

  $\begin{array}{l}{a}_1=18,a_n=a_{n-1}-4,n\ge 2\\ \text{Put }n=2\\ a_2=a_{2-1}-4\\ a_2=a_1-4\\ \text{As }{a}_1=18\\ a_2=18-4=14\\ \text{Now put }n=3\text{ in }{a}_n=a_{n-1}-4,n\ge 2\\ a_3=a_{3-1}-4\\ a_3=a_2-4\\ \text{As }{a}_2=14\\ a_3=14-4=10\\ \text{Now put }n=4\text{ in }{a}_n=a_{n-1}-4,n\ge 2\\ a_4=a_{4-1}-4\\ a_4=a_3-4\\ \text{As }{a}_3=10\\ a_4=10-4=6\\ \text{Now put }n=5\text{ in }{a}_n=a_{n-1}-4,n\ge 2\\ a_5=a_{5-1}-4\\ a_5=a_4-4\\ \text{As }{a}_4=6\\ a_5=6-4=2\end{array}$

Conclusion:

First five terms of the sequence are $a_1=18,a_2=14,a_3=10,a_4=6\&a_5=2$