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# Help???

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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
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#1
+76094
+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 / 2   =

12

CPhill  Sep 11, 2017
#2
+332
+1

Thanks CPhill

MIRB16  Sep 11, 2017
#3
+76094
+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

CPhill  Sep 11, 2017

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