Chapter 7.8, Problem 3P

a.

To determine

To write: a recursive formula and an explicit formula for the given sequence.

Answer to Problem 3P

  $g(x)=g(x-1)\times 6$

  $g(x)=14\times {(6)}^{x-1}$

  $g(8)=3919104$

Explanation of Solution

Given information: a sequence is given as follows-

14, 84, 504, 3024 . . .

Here the ratio $\frac{84}{14}=\frac{504}{84}=\frac{3024}{504}=6$ is same

So, the given sequence is in geometric progression whose common ratio is 6 and first term is 14.

Then $x^{th}$ term is given by $g(x)=14\times {(6)}^{x-1}$

Now every term is obtained by multiplying by 6 in the preceding term, so the recursive formula of given geometric sequence is given by $g(x)=g(x-1)\times 6$ .

And the $8^{th}$ term of sequence is, $g(8)=14{(6)}^{8-1}=14\times {(6)}^7=3919104$ .

b.

To determine

To write: a recursive formula and an explicit formula for the given sequence.

Answer to Problem 3P

  $\begin{array}{l}g(x)=g(x-1)\times \frac{1}{2}\\ g(x)=648\times {\left(\frac{1}{2}\right)}^{x-1}\\ g(8)=\frac{81}{16}\end{array}$

Explanation of Solution

Given information: a sequence is given as follows-

648, 324, 162, 81 . . .

Here the ratio $\frac{648}{324}=\frac{324}{162}=\frac{81}{162}=\frac{1}{2}$ is same

So, the given sequence is in geometric progression whose common ratio is $\frac{1}{2}$ and first term is 648.

Then $x^{th}$ term is given by $g(x)=648\times {\left(\frac{1}{2}\right)}^{x-1}$

Now every term is obtained by multiplying by $\frac{1}{2}$ in the preceding term, so the recursive formula of given geometric sequence is given by $g(x)=g(x-1)\times \frac{1}{2}$ .

And the $8^{th}$ term of sequence is $g(8)=648{\left(\frac{1}{2}\right)}^{8-1}=648\times {\left(\frac{1}{2}\right)}^7=\frac{81}{16}$ .