Question

The fourth term of a sequence is 108. Each term after the first is 3 times the previous term. Write an explicit expression that models the general term of the sequence f(n). Write your answer in the space provided on your answer document.

Answer 1

Answer:

$f(n)=4(3)^{n-1}$

Step-by-step explanation:

Let the first term be $x$, then the terms of the sequence are:

$x,3x,9x,27x,81x,...$.

Since the fourth term is 108, we have

$27x=108$

$\implies x=4$

Hence the sequence becomes:

$4,12,36,108,324,...$.

The explicit expression for a geometric expression is given by:

$f(n)=ar^{n-1}$

where a=4 and r=3

The required formula is:

$f(n)=4(3)^{n-1}$

Answer 2

The general term of the sequence is f(n) = 4 * 3^(n-1)

How to determine the general term?

The given paramaters are:

f(4) = 108

Common ratio (r) = 3

The nth term of a geometric progression is:

f(n) = ar^(n-1)

So, we have:

f(4) = a * 3^(4-1)

Substitute 108 for f(4)

108= a * 3^3

Evaluate the quotient

108= a * 27

Divide both sides by 27

a = 4

Recall that:

f(n) = ar^(n-1)

So, we have:

f(n) = 4 * 3^(n-1)

Hence the general term of the sequence is f(n) = 4 * 3^(n-1)

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