Question

What is the 6th term of the geometric sequence where a1 = 1,024 and a4 = −16? 1 −0.25 −1 0.25

Answer 1

The n-th term is given by

$a_n=a_1\cdot r^{(n-1)}\qquad\text{where r is the common ratio}$

Then we can find the common ratio from the given terms.

$\dfrac{a_4}{a_1}=\dfrac{a_1\cdot r^{(4-1)}}{a_1}=r^3=\dfrac{-16}{1024}=\left(\dfrac{-1}{4}\right)^3\\\\r=\dfrac{-1}{4}\\\\a_6=1024\left(\dfrac{-1}{4}\right)^5=-1$

The appropriate choice is -1.

Answer 2

Answer:

Option 3rd is correct

$a_6 = -1$

Step-by-step explanation:

The nth term for the geometric sequence is given by:

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

where,

$a_1$ is the first term

r is the common ratio

n is the number of terms.

As per the statement:

$a_1 = 1024$

$a_4 = -16$

For n = 4, we have;

$a_4 = a_1 \cdot r^3$

⇒$-16 = 1024 \cdot r^3$

Divide both sides by 1024 we have;

$-\frac{1}{64} =r^3$

⇒$r =\sqrt[3]{-\frac{1}{64}}=\sqrt[3]{-\frac{1}{4^3}}$

⇒$r = -\frac{1}{4} = -0.25$

We have to find the value of  6th term.

for n = 6

$a_6 = 1024 \cdot (-0.25)^5 = 1024 \cdot (-0.0009765625) = -1$

⇒$a_6 = -1$

Therefore,  the 6th term of the geometric sequence is, -1