Question

A geometric sequence has terms a3 = 2 and a9 = 128. Find the explicit formula for this sequence.

Answer 1

we know that

the explicit formula in a geometric sequence is

$a_n=a_1\cdot(r)^{(n-1)}$

In this problem

we have that

a3=2

a9=128

substitute in the expression above

$\begin{gathered} a_3=a_1\cdot(r)^{(3-1)} \\ 2=a_1\cdot r^2 \end{gathered}$

and

$\begin{gathered} a_9=a_1\cdot(r)^{(9-1)} \\ 128=a_1\cdot r^8 \end{gathered}$

Divide both expressions

128=a1*(r^8)

2=a1*(r^2)

----------------

128/2=r^8/r^2

64=r^6

r=2

Find out the value of a1

2=a1*(r^2)

2=a1*(2^2)------> 2=a1*4

a1=1/2

therefore

a1=1/2 and r=2

the explicit formula is equal to

$a_n=\frac{1}{2}\cdot(2)^{(n-1)}$

the answer is option C