(98 ) can someone take a look at this? i'm stumped. (answer choices provided)

what is the fifth term of the geometric sequence?

geometric sequence represented by a graph in quadrant 1 with n on x axis and a sub n on y axis. points at (1,81), (2,9), and (3,1). (graph attached)

hint: an = a1(r)n − 1, where a1 is the first term and r is the common ratio

a5 = 1/81

Step-by-step explanation:

(1,81) (2,9) (3,1)

The sequence goes

81,9,1,

We are multiplying by 1/9 each time

a1 = 81

r= 1/9

an = a1 r^(n-1)

an = 81 * (1/9) ^ (n-1)

The 5th terms

a5 = 81 * (1/9) ^ (5-1)

a5 = 81 * (1/9) ^4

a5 = 81/6561

a5 = 1/81

Step-by-step explanation:

(1, 81), (2, 9), (3, 1) means

the common ratio (r) is: 9 ÷ 81  →  r =  1/9

therefore,

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.