Question

Which recursive formula can be used to generate the sequence shown, where f(1) = 9.6 and n > 1?

Answer 1

The correct answer is f(n + 1) = –0.5f(n) since n is greater than 1

Answer 2

Answer:

$f(n)=-0.5\times f(n-1)$

Step-by-step explanation:

Sequence: 9.6, –4.8 , 2.4, –1.2, 0.6, ...

So, f(1) = first term = 9.6

r = common ratio = $\frac{-4.8}{9.6} =\frac{2.4}{-4.8} = -0.5$

Now , formula of nth term in G.P. = $f(n)=f(1)\times r^{n-1}$

So, formula for nth term of the given sequence =  $f(n)=9.6\times (-0.5)^{n-1}$

So,  $f(n-1)=9.6\times (-0.5)^{n-1-1}$

      $f(n-1)=9.6\times (-0.5)^{n-2}$

Recursive formula :

$\frac{f(n)}{f(n-1)}= \frac{9.6\times (-0.5)^{n-1}}{9.6\times (-0.5)^{n-2}}$

$\frac{f(n)}{f(n-1)}=(-0.5)^{n-1-(n-2)}$

$\frac{f(n)}{f(n-1)}=(-0.5)^{-1+2}$

$\frac{f(n)}{f(n-1)}=-0.5$

$f(n)=-0.5\times f(n-1)$

Hence recursive formula can be used to generate the sequence shown, where f(1) = 9.6 and n > 1 is $f(n)=-0.5\times f(n-1)$