algebra and trigonometry

Question

the sixth term of a geometric series of positive number is 10 and the sixteenth term is 0.1. What is the eleventh term? Find its sum to infinity.

Answer 1

a6=a1*(r^5)=10

 

a16=a1*(r^15)=0.1

 

Solve the above two equations for a1. &. r

 

r =0.6309

 

a1=100

 

We can check these values by plugging in back in the above equations.

 

Eleventh term-----

 

a11=a1*(r^10)=100(0.6309^10)

                       =0.999

 

Sum of terms up to infinite ---

 

S= a1/(1-r)

 

   =100/0.3691

 

   =271 

Answer 2

the i'th term of a geometric series is

 

(1)     a_i=a₀r^(i-1)

 

You are told that

 

(2)     a₆=a₀r⁵=10

 

(3)     a₁₆=a₀r¹⁵=0.1

 

Divide (3) by (2)

 

(4)     r¹⁰=(0.1)/10=0.01=1/100

 

Take the tenth root

 

(5)     r=(0.01)^(1/10)=0.69095

 

Find a₀ by substituting from (5) into (2)

 

(6)      a₆=a₀(0.69095)⁵=a₀(0.1)=10

 

(7)     a₀=10/0.1=100

 

The eleventh term is given by

 

(8)     a₁₁=a₀r¹⁰=100(1/100)=1

 

The formula for the sum of an infinite geometric series is

 

S=a0/(1-r)=1/(1-0.69095)=2.71

 

Check the algebra and computations for errors.