Question

Identify the 7th term of the geometric sequence in which a2 = 324 and a4 = 36

Answer 1

The answer is 4/3 or -4/3

Answer 2

Answer:

4/3

Step-by-step explanation:

a2 = 324 and a4 = 36

second term is 324  and fourth term is 36

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

To get second term, replace n with 2

$a_2= a_1(r)^{2-1}$

$a_2= a_1(r)$

replace a2 with 324

$324= a_1(r)$ -----------> equation 1

To get fourth term , replace n with 4

$a_4= a_1(r)^{4-1}$

$a_4= a_1(r)^3$      

Replace the value

$36= a_1(r)$  -----------> equation 2

divide second equation by first equation

$36= a_1(r)^3$    divide by $324= a_1(r)$

1/9 = r^2

take square root on both sides

1/3= r

now we find a1

$324= a_1(r)$  

$324= a_1(\frac{1}{3})$

multiply by 3 on both sides

972= a_1

So a1= 972,  r= 1/3

now find out 7 term by replacing 'n' with 7

$a_7= 972(\frac{1}{3})^{7-1}$

So 7th term is 4/3