Question

The third term in a geometric sequence is -81. The common ratio is 1/3

Write a exponential equation for this sequence.

Answer 1

`t(n)=-729((1)/(3)) ^(n-1)`

Explanation:

Given:

A geometric sequence with third term = -81

Common ratio =
`(1)/(3)`

General term of a geometric sequence is given by the formula:

`t(n)=ar^(n-1)`

where :

t(n) is nth term

a is the first term

r is the common ratio

n=1,2,3,4...

Here r=1/3 and t(3)=-81

`ar^(2)=-81\\a*(1)/(3)^(2)=-81\\a=-729`

General equation becomes:

`t(n)=-729((1)/(3)) ^(n-1)`

Answer 2

Explanation:

A geometric sequence is a sequence in which the successive terms increase or decrease by a common ratio. The formula for the nth term of a geometric sequence is expressed as follows

Tn = ar^(n - 1)

Where

Tn represents the value of the nth term of the sequence

a represents the first term of the sequence.

n represents the number of terms.

From the information given,

r = 1/3

T3 = - 81

n = 3

Therefore,

- 81 = a× 1/3^(3 - 1)

-81 = a × (1/3)^2

-81 = a/9

a = -81 × 9 = - 729

The exponential equation for this sequence is written as

Tn = - 729 * (1/3)^(n-1)