Equation woed problem

Question

Set up the equation for the 22nd term of the geometric sequence in which the 4th term is 3 and the 10th term is 6.

Answer 1

a_n = (a₁)r^(n-1)

 

a₃ = 3 = (a₁)r^(4-1)

(a₁)r³ = 3

 

and

 

a₁₀ = 6 = (a₁)r^(10-1)

(a₁)r⁹ = 6

 

You now have a system of exponential equations. Solve both for a₁ in terms of r and find r. Then use r to find a₁.

 

a₁ = 3/r³   and    a₁ = 6/r⁹

 

According to the transitive property, two things equal to the same thing equal each other:

 

3/r³ = 6/r⁹

 

Cross-multiply and rearrange algebraically

 

3r⁹ = 6r³

 

r⁹/r³ = 6/3

 

r⁶ = 2

 

r= 2^(1/6)

 

Now we can use either equation to get a₁.

 

a₁ = 3/r³ = 3/2^(3/6) = 3/2^(1/2)

 

 

a₂₂ =[3/2^(1/2)]*[2^(1/6)]²¹ = [3/2^(1/2)]*[2^(21/6)] = 3/2^(1/2) × 2^(7/2) = 3 × 2³ = 24