Question

What is the explicit rule for this geometric sequence? a1=-15;an=1/5•an-1

Answer 1

Answer:

Option A is correct.

$a_n =-15 (\frac{1}{5})^{n-1}$

Step-by-step explanation:

Given: $a_1 = -15$ ; $a_n = \frac{1}{5} \cdot a_{n-1}$

for n = 2

$a_2 = \frac{1}{5} \cdot a_{2-1}$ = $\frac{1}{5}\cdot a_1 = \frac{1}{5} \cdot (-15)$

for n =3

$a_3 = \frac{1}{5} \cdot a_{3-1}$ = $\frac{1}{5} \cdota_2= (\frac{1}{5})^2\cdot (-15)$

Simlarly , for n =4

$a_3 = \frac{1}{5} \cdot a_{4-1}$ = $\frac{1}{5}\cdot a_3= (\frac{1}{5})^3\cdot (-15)$

and so on...

Common ratio(r) states that for a geometric sequence or geometric series, the common ratio is the ratio of a term to the previous term.

we have;   $r = \frac{a_2}{a_1} = \frac{1}{5}$

Now; by recursive formula for geometric series:

$a_n = a_1 r^{n-1}$

where $a_1$ is the first term and r is the common ratio:

Substitute the value of $a_1 = -15$  and  $r = \frac{1}{5}$

we have;

$a_n =-15 (\frac{1}{5})^{n-1}$

therefore, the explicit rule for the geometric sequence is; $a_n =-15 (\frac{1}{5})^{n-1}$