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# Algebra Help

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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
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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
+75394
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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......

CPhill  Feb 18, 2017
#3
+75394
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

CPhill  Feb 18, 2017

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