Chapter 7.10, Problem 40PFA

To determine

The sixth term of the given sequence.

Answer to Problem 40PFA

The sixth term is -2389.

Explanation of Solution

Given information: A table given with n and $a_n$ values. When n=1, $a_n=-2$ , when n=2, $a_n=-9$ , when n=3, $a_n=-37$ , when n=4, $a_n=-149$ .

Calculation:

Here given, the first term $a_1=-2$ , second term $a_2=-9$ , third term $a_3=-37$ , fourth term $a_4=-149$ .

So, it can be written as,

  $\begin{array}{l}{a}_2=-9=-2\times 4-1=a_1\times 4-1=4a_1-1\\ a_3=-37=-9\times 4-1=a_2\times 4-1=4a_2-1\\ a_4=-149=-37\times 4-1=a_3\times 4-1=4a_3-1\end{array}$

  $\therefore$ The recursive formula is $a_n=4a_{n-1}-1$ .

  $\therefore$ Fifth term

  $\begin{array}{l}{a}_5=4a_4-1\\ \Rightarrow a_5=4(-149)-1\\ \Rightarrow a_5=-597\end{array}$

  $\therefore$ Sixth term

  $\begin{array}{l}{a}_6=4a_5-1\\ \Rightarrow a_6=4(-597)-1\\ \Rightarrow a_6=-2389\end{array}$

So, the sixth term of the given sequence is -2389.

Option A is correct.