Chapter 4.7, Problem 62PPE

To determine

To find : the second, third and fourth terms of the sequence and the explicit formula to represent the sequence whose recursive formula is $A\left(n\right)=A\left(n-1\right)+3;A\left(1\right)=13$ .

Answer to Problem 62PPE

The second, third and fourth terms of the sequence are 16, 19and 22 respectively.

The explicit formula for given sequence is

  $A\left(n\right)=13+\left(n-1\right)\left(3\right)$

Explanation of Solution

Given information: The recursive formula for the sequence is $A\left(n\right)=A\left(n-1\right)+3;A\left(1\right)=13$

Calculation:

By substituting n= 2,3 and 4 and then by using the previous term, we get the second, third and fourth terms of the sequence.

  $\begin{array}{l}A\left(2\right)=A\left(2-1\right)+3=A\left(1\right)+3=13+3=16\\ A\left(3\right)=A\left(3-1\right)+3=A\left(2\right)+3=16+3=19\\ A\left(4\right)=A\left(4-1\right)+3=A\left(3\right)+3=19+3=22\end{array}$

Thus, the second, third and fourth terms of the sequence are 16, 19and 22 respectively.

Now, the first term and common difference of the sequence is $A\left(1\right)=13$ and $d=3$ respectively.

Now, the explicit formula for the arithmetic sequence with first term $A\left(1\right)$ and common ratio d is

  $A\left(n\right)=A\left(1\right)+\left(n-1\right)\left(d\right)$

Thus, explicit formula for given sequence is

  $A\left(n\right)=13+\left(n-1\right)\left(3\right)$