Chapter 8.5, Problem 51E

To determine

To find: the 1000th term of the sequence whose first term and nth term is given.

Answer to Problem 51E

B. $a_{1000}=5998$ .

Explanation of Solution

Given:

The nth term of given sequence is $a_n=a_{n-1}+6$ .

  $a_1=4$

The given options are:

1. 4006
2. 5998
3. 1010
4. 10,000

Calculation:

The given sequence is $a_n=a_{n-1}+6$ .

Therefore, $d=6$

Now, according to the question of $a_1=4$ . Therefore,

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)d\\ \Rightarrow a_n=4+\left(n-1\right)\left(6\right)\left\{\text{substitutes these value}\right\}\\ \therefore a_n=4+6n-6\\ \Rightarrow a_n=6n-2\end{array}$

Therefore, the 1000th term of the sequence is

  $\begin{array}{l}{a}_{1000}=6\left(1000\right)-2\\ a_{1000}=6000-2=5998\end{array}$

Hence, the recursive rule for the given sequence is

  $a_{1000}=5998$ .

Therefore, the correct option is (B).