find the 8th term of a geometric sequence whose second term is 6 and fifth term is 81/4. write your final answer as a fraction

Answer 1

g_n=g₁r^n-1

g₂=g₁r=6 --> g₁=6/r

g₅=(6/r)r⁴ = 6r³=81/4 --> r = (81/24)^1/3=(27/8)^1/3 = 3/2

g₁=6/r=6/(3/2)=4

g₈=4(3/2)⁷=4(2187)/128 = 2187/32

Answer 2

The explicit formula to get a term in geometric sequence is:

a_n = a₁•r^n-1

Given:

a₂ =6

a₅ = 81/4

a₈ = ? (which is what we are looking for)

Base on the formula we can have the following:

a₂ = a₁•r^2-1 = a₁•r = 6

a₅ = a₁•r^5-1= a₁•r⁴= 81/4

There are two equations and two unknowns, therefore this is systems of equation

From the first equation, we can have

a₁= 6/r

Then substitute a₁ in the second equation:

(6/r)•r⁴= 81/4

6r³= 81/4

multiply 1/6 on both sides:

r³= 27/8

get the cube root of both sides:

r = 3/2

For a₁ use a₁= 6/r:

a₁ = 6/(3/2) = 6•(2/3) = 4

To find the 8th term of the sequence, use the formula and then substitute the value of a₁ and r:

a₈ = a₁•r⁷

a₈ = (4)(3/2)⁷

a₈ = (4)(2187/128)

a₈ = 2187/32

Answer 3

a2 = 6

a5 = 81/4

a8=?

3d = 81/4 - 6 = (81-24)/4 =57/4

d = 19/4

a1 = 6-19/4 = 5/4

an=a1 +(n-1)d

a8 = 5/4 +7(19/4) = (5+ 133)/4 = 138/4 = 69/2

a1 = 5/4

a2 = 6

a3 = 43/4

a4 = 62/4

a5=81/4

a6 =81/4 + 19/4= 100/4

a7 = 100/4 + 19/4 = 119/4

a8 = 119/4 + 19/4 = 138/4 =69/2

a8 = 81/4 + 3(19/4) = (81+57)/4 = 138/4 = 69/2