Chapter 9, Problem 29RE

Solution

To state: The first six terms and the 12^th term of the sequence:

  $a_1=-1\text{and}{a}_n=a_{n-1}+3\text{for}n\ge 2$ .

The resultants first six terms are $-1,2,5,8,11,14$ and the 12^th term is 32.

Given information:

The given sequence is: $a_1=-1\text{and}{a}_n=a_{n-1}+3\text{for}n\ge 2$ .

Explanation:

Consider the given sequence,

  $a_n=a_{n-1}+3$

The first term of the sequence is: $a_1=-1$

Now substitute 2 for n for the second term and simplify,

  $\begin{array}{c}{a}_2=a_{2-1}+3\\ =a_1+3\\ =-1+3\\ =2\end{array}$

Now substitute 3 for n for the third term and simplify,

  $\begin{array}{c}{a}_3=a_{3-1}+3\\ =a_2+3\\ =2+3\\ =5\end{array}$

Similarly, find the remaining 3 terms,

  $\begin{array}{c}{a}_4=a_{4-1}+3\\ =a_3+3\\ =5+3\\ =8\\ a_5=a_{5-1}+3\\ =a_4+3\\ =8+3\\ =11\\ a_6=a_{6-1}+3\\ =a_5+3\\ =11+3\\ =14\end{array}$

Similarly, the 12^th term is:

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ a_{12}=-1+\left(12-1\right)3\\ =-1+11\cdot 3\\ =-1+33\\ =32\end{array}$

Therefore, the first six terms are $-1,2,5,8,11,14$ and the 12^th term is 32.