Question

In Problems 37–50, a sequence is defined recursively. Write down the first five terms.

$a_1=3;a_n=4-a_{n-1}$

Short answer

The first five terms of the recursively defined sequence are$3,1,3,1,\mathrm{and}3$.

Step-by-step solution

Step 1. Write the given information.

The given recursively defined sequence is:

$$\begin{gathered} a_1=3 \\ {a}_n=4-a_{n-1} \end{gathered}$$

Step 2. Determine the first term and find out the second term.

The first term is $\left(a_1\right)=3$.

Now substitute 2 for nin the given formula $a_n=4-a_{n-1}$to get the second term,

$$\begin{gathered} a_2=4-a_{2-1} \\ =4-a_1 \\ =4-3 \\ =1 \end{gathered}$$

Step 3. Find the 3rd and 4th terms.

Similarly substitute 3, 4 for nin the given formula to get the third and fourth term,

$$\begin{gathered} a_3=4-a_{3-1} \\ =4-a_2 \\ =4-1 \\ =3 \\ {a}_4=4-a_{4-1} \\ =4-a_3 \\ =4-3 \\ =1 \end{gathered}$$

Step 4. Find the 5th term.

Now substitute 5 for nin the given formula to get the 5th term,

$$\begin{gathered} a_5=4-a_{5-1} \\ =4-a_4 \\ =4-1 \\ =3 \end{gathered}$$

Therefore, the first five terms are$3,1,3,1,3$.