Answer :
We begin by writing the sequence:
[tex]$$
a_1 = 44,\quad a_2 = -11,\quad a_3 = \frac{11}{4},\quad a_4 = -\frac{11}{16},\quad \ldots
$$[/tex]
Step 1: Calculate the ratio between the second term and the first term:
[tex]$$
\text{ratio} = \frac{a_2}{a_1} = \frac{-11}{44} = -\frac{1}{4}.
$$[/tex]
Step 2: Verify that this common ratio is consistent by computing the third term:
[tex]$$
a_3 = a_2 \times \left(-\frac{1}{4}\right) = -11 \times \left(-\frac{1}{4}\right) = \frac{11}{4}.
$$[/tex]
Since the ratio remains [tex]$-\frac{1}{4}$[/tex] from term to term, the sequence is geometric.
Step 3: Therefore, the recursive formula for the sequence is:
[tex]$$
f(n) = -\frac{1}{4} \, f(n-1) \quad \text{for } n > 1.
$$[/tex]
Among the options provided, this corresponds to the third option.
[tex]$$
a_1 = 44,\quad a_2 = -11,\quad a_3 = \frac{11}{4},\quad a_4 = -\frac{11}{16},\quad \ldots
$$[/tex]
Step 1: Calculate the ratio between the second term and the first term:
[tex]$$
\text{ratio} = \frac{a_2}{a_1} = \frac{-11}{44} = -\frac{1}{4}.
$$[/tex]
Step 2: Verify that this common ratio is consistent by computing the third term:
[tex]$$
a_3 = a_2 \times \left(-\frac{1}{4}\right) = -11 \times \left(-\frac{1}{4}\right) = \frac{11}{4}.
$$[/tex]
Since the ratio remains [tex]$-\frac{1}{4}$[/tex] from term to term, the sequence is geometric.
Step 3: Therefore, the recursive formula for the sequence is:
[tex]$$
f(n) = -\frac{1}{4} \, f(n-1) \quad \text{for } n > 1.
$$[/tex]
Among the options provided, this corresponds to the third option.
