Question

Given the functions f(n) = 11 and g(n) = (three fourths)n − 1, combine them to create a geometric sequence, an, and solve for the 9th term.
an = (11 • three fourths)n − 1; a9 ≈ 24.301
an = 11(three fourths)n − 1; a9 ≈ 1.101
an = 11 + (three fourths)n − 1; a9 ≈ 11.100
an = 11 − (three fourths)n − 1; a9 ≈ 9.900

Answer 1

We are given two functions:
f(n) = 11
g(n) = (3/4) ^(n-1)
I have rewritten the functions to their correct form. Notice that the term (n - 1) is the exponent of 3/4.

We are asked to combine the two functions to model a geometric sequence and solve for the 9th term.

The general formula for a geometric sequence is
an  = a1 r^(n - 1)

From the given functions, we can set
f(n) = a1 = 11
and 
g(n) = r^(n - 1) = (3/4)^(n - 1)

Substituting to the general formula of a geometric sequence, the result is
an = 11 (3/4)^(n - 1)

Solving for the 9th term
a9 = 11 (3/4)^(9 - 1)
a9 = 1.101

The answer is the second option.

Answer 2

Answer:

Option 2

$a_n=11\times \frac{3}{4}^{n-1}; a_9=1.101$

Step-by-step explanation:

Given : The functions $f(n) = 11$ and $g(n) = \frac{3}{4}^{n-1}$ combine them to create a geometric sequence, an

To find : The 9th term

Solution :

The two function are :

$f(n) = 11$

$g(n) = \frac{3}{4}^{n-1}$

We have to form a geometric sequence by combining f(n) and g(n).

So, the sequence having nth term is

$a_n=f(n)\times g(n)$

$a_n=11\times \frac{3}{4}^{n-1}$  ........[1]

Now, we put n=1,2,3,... to make a sequence

Put n=1 in [1]

$a_2=11\times \frac{3}{4}^{2-1}$

$a_2=11\times \frac{3}{4}$

$\text{Common ratio}=\frac{\text{Second Term}}{\text{First Term}}$

$r=\frac{a_2}{a_1}$

$r=\frac{11\times \frac{3}{4}}{11}$

$r=\frac{3}{4}$

The formula of nth term in geometric sequence is

$T_n=a_1\times r^{n-1}$

Put n= 9 to find 9th term

$T_9=11\times \frac{3}{4}^{9-1}$

$T_9=11\times \frac{3}{4}^{8}$            

$T_9=11\times 0.1001$

$T_9=1.101$

Therefore, Option 2 is correct.

$a_n=11\times \frac{3}{4}^{n-1}; a_9=1.101$