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**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**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......

CPhill
Feb 18, 2017

#3**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