+0  
 
0
693
3
avatar

Help for these two problems:

 

 

 

In an infinite geometric progression with positive terms and with a common ratio |r|<1, the sum of the first three terms is 

(26/3) and the sum of the entire progression is 9. Determine the progression.

 

 

 

 

 

Find the sum of the first five terms of an infinite geometric progression with a common ratio |r|<1 if the second term is (4/3) and the ratio of the sum of the squares of the terms of the progression to the sum of the terms of the progression is 3:1.

Guest Feb 18, 2017
 #1
avatar
0

Find the sum of the first five terms of an infinite geometric progression with a common ratio |r|<1 if the second term is (4/3) and the ratio of the sum of the squares of the terms of the progression to the sum of the terms of the progression is 3:1.

 

Progression = 4, 4/3, 4/9, 4/27, 4/81..........

Common ratio =1/3

Sum(5) =4 / [1 - 1/3] = 6

The terms of the progression converge to 6

The sum of the squares of the terms converge to 18.

 

 

In an infinite geometric progression with positive terms and with a common ratio |r|<1, the sum of the first three terms is 

(26/3) and the sum of the entire progression is 9. Determine the progression.

 

Progression=6, 2, 2/3, 2/9, 2/27, 2/81............

Common ratio =1/3

Sum of the first 3 terms =26/3

Sum of entire progression = 9

Guest Feb 18, 2017
 #2
avatar+87301 
0

In an infinite geometric progression with positive terms and with a common ratio |r|<1, the sum of the first three terms is 

(26/3) and the sum of the entire progression is 9. Determine the progression.

 

We have that the sum of the infinite series is given by :

 

a /(1 - r)  = 9      where a is the first term and  r  = the common ratio between terms.....solve for a.....

 

a = 9(1 - r)

 

For the sum of the first three terms, we have :

 

26/3  = a(1 - r^3)/ (1 - r)         sub for  "a"

 

26/3  = 9(1 -r) * (1 - r^3)  / ( 1 - r)

 

26/3 = 9 (1 - r^3 )

 

26/27  = 1  - r^3

 

r^3 =  1 - 26/27   =  1/27      take the cube root of both sides

 

So

 

r = 1/3

 

And a  =  9 (1 -r)  = 9(1 - 1/3)  = 9 * 2/3   =  6

 

And the progression is

 

6 , 2 , 2/3 , 2/9 , 2/27......

 

 

 

cool cool cool

CPhill  Feb 18, 2017
 #3
avatar+87301 
0

Find the sum of the first five terms of an infinite geometric progression with a common ratio |r|<1 if the second term is (4/3) and the ratio of the sum of the squares of the terms of the progression to the sum of the terms of the progression is 3:1.

 

Call the first term  "a"    so....the second term  is 

 

ar  =4/3  →    a  = 4/[3r]

 

And we know that

 

[ a^2 + (ar)^2 + (ar^2)^2 +  ...... + (ar^n)^2 ]  /   [a + ar + ar^2 +......+  ar^n]  = 3

 

The sum of the series in the denominator can be written as  :   a^2 / [1 - r^2]

 

And the sum of the series in the numerator  can be written as :    a / [ 1 - r]

 

And we have that

 

(  a^2 / [1 - r^2] ) / ( a / [ 1 - r] )  = 3   simplify

 

a(1 -r) / (1 - r^2)  = 3

 

a(1 - r) / [ (1 - r) (1 + r)]  = 3

 

a/(1 + r)  = 3

 

a = 3(1 + r)          sub for "a"

 

4/[3r]  = 3(1 + r)

 

4 = 9r(1 + r)

 

4 = 9r + 9r^2      rearrange

 

9r^2 + 9r - 4  = 0  factor

 

(3r + 4) (3r - 1)  = 0    set each factor to 0   and  r = 1/3  or r = -4/3

 

But ....  l -4/3 l   >  1   so reject this solution

 

So r  = 1/3     and  a  = 4/ [ 3 *1/3]  = 4

 

And the sum of the series =   4 / [1 - 1/3] =  4 / (2/3)  = 12 / 2  = 6

 

And the sum  of the squares of the series =  

 

 4^2 / [1 - (1/3)^2] =   16 / [ 8/9]  =  16 * 9 / 8  =  (16/8) * 9  =  2 * 9   = 18

 

And   18 / 6    =  3

 

And the sum of the first 5 terms =  

 

4 [1 - (1/3)^5] / (1 - 1/3)  =   4 [ 1 - 1/243] / [2/3] =  6 [ 242/243]  = 2*242 / 81  =  484 / 81

 

 

cool cool cool

CPhill  Feb 18, 2017

11 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.