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The geometric series \(a+ar+ar^2+\cdots\) has a sum of 12, and the terms involving odd powers of r have a sum of 5. What is r?

MIRB16  Sep 11, 2017
 #1
avatar+91071 
+3

 

The sum of the series  can be represented by :

 

a / ( 1 - r)  = 12

 

a = 12 ( 1- r )    (1)

 

So....note that the sum of the  terms  with odd powers of r  can be represented by :

 

ar  + ar^3  + ar^5 + ar^7 +  ..... + ar^(2n - 1)  =  5   (2)

 

And note that the sum of the  terms  with even powers of r  can be represented by :

 

a  +  ar^2  + ar^4 + ar^6 + ar^8 +  ..... + ar^(2n)  =  7   (3)

 

Multiply (2)  by r on both sides

 

ar^2  + ar^4  + ar^6 + ar^8 +  ..... + ar^(2n )  =  5r     (4)

 

Subtract   (4)  from (3)

 

a  = 7 - 5r    (5)

 

Sub (5)  into (1)

 

7 - 5r  = 12(1 - r)

 

7 - 5r  =  12 - 12r

 

7r = 5     →   r  =  5/7

 

Check

 

a = 7 - 5(5/7)  =  24/7

 

a / ( 1 - r)  =

 

(24/7) / ( 1- 5/7)  =

 

(24/7) / ( 2/7)  =

 

24 / 2   =

 

12 

 

 

cool cool cool

CPhill  Sep 11, 2017
 #2
avatar+738 
+4

Thanks CPhill

MIRB16  Sep 11, 2017
 #3
avatar+91071 
+1

Here's another way of attacking this......

 

Note that  the sum of the  terms  with odd powers of r  can be represented by :

 

ar  + ar^3  + ar^5 + ar^7 +  ..... + ar^(2n - 1)  =  5   

 

So.....the common ratio between these terms is just r^2

 

And the sum of this series can be represented as

 

5  =   ar / ( 1 - r^2)   →   (1 - r^2)  =  ar / 5     (1)

 

And note that the sum of the  terms  with even powers of r  can be represented by :

 

a  +  ar^2  + ar^4 + ar^6 + ar^8 +  ..... + ar^(2n)  =  7

 

And the common ratio between these terms is just r^2

 

And the sum of this series can be represented by

 

7  = a / (1 - r^2)  →   (1 - r^2)  =  a / 7    (2)

 

Equating (1)  and (2)  we have that

 

ar / 5   =  a / 7              divide both sides by a

 

r / 5   =  1 / 7               multiply both sides by 5

 

r =  5 / 7

 

 

cool cool cool

CPhill  Sep 11, 2017

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