Question

Find the eighth term of the following geometric sequence 14,-7, 7/2

Answer 1

You can see that you get the next term by dividing by 2 and switching sign (that is, dividing by -2).

So, the general term is

$a_n = \dfrac{14}{(-2)^{n-1}}$

In fact, you can check that

$a_1 = \dfrac{14}{(-2)^0}=\dfrac{14}{1}=14$

$a_2=\dfrac{14}{(-2)^1}=\dfrac{14}{-2}=-7$

$a_3=\dfrac{14}{(-2)^2}=\dfrac{14}{4}=\dfrac{7}{2}$

So, the eight term is

$a_8=\dfrac{14}{(-2)^7}=\dfrac{7}{-2^6}=-\dfrac{7}{64}$

Answer 2

7/32
Keep the pattern going. A(n)=A(n-1)*-1/2
14, -7, 7/2, -7/4, 7/8, -7/16, 7/32