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# In a geometric sequence, t4=5x-11, t5=x+5 and t6=x-1. Find the sum(s) of the infinite series if it (or they) exist(s).

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In a geometric sequence, t4=5x-11, t5=x+5 and t6=x-1. Find the sum(s) of the infinite series if it (or they) exist(s).

So uh... what does that mean and what do I do and how do I do it? I really have no clue how to even start answering this question...

May 29, 2017

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In order for this to be geometric, we have to have that

[ 5 x - 11] * r   =  x + 5        and that

[ x + 5 ] * r   =  x - 1          where r is the common ratio

Manipulating the first equation by dividing both sides by [ 5x - 11], we have that

r  =  [ x + 5 ] / [ 5x - 11]

Substituting for r in the second equation, we have

[x + 5] [x + 5] / [ 5x - 11]  =  x - 1        multiply both sides by 5x - 11, simplify the right side

x^2 + 10x + 25   =  [ 5x - 11] [ x - 1]     simplify the right side

x^2 + 10x + 25   =  5x^2 - 11x - 5x  + 11

x^2 + 10x + 25  =  5x^2 - 16x + 11      subtract  x^2 + 10x + 25 from both sides

4x^2 - 26x - 14  =  0     divide through  by  2

2x^2  - 13x - 7  =  0     factor

(2x  + 1) ( x - 7)   =  0

Setting both factors to 0 and solving for x we have that  x  = -1/2  or x  = 7

So.....if x  = -1/2, then r  =  [ -1/2+ 5 ] / [ 5(-1/2) - 11]  =  [ 4.5] / [-13.5] =   (-1/3)

And the  4th, 5th and 6th terms  are    =  -27/2, 9/2, -3/2

And....the first term of this series  =  (-27/2) (-3)^3  =  729/2

And the sum of this infinite series  =   [first term] / [ 1 - r ]  =

[ 792/2] /  [ 1  - (-1/3) ]  = 396/ (4/3)  = 396 * (3/4) =  297

And   when x   =  7,  r   =  [ 7 + 5] / [ 5*7 - 11]  =  12/ 24  =  1/2

And the 4th, 5th and 6th terms are  24, 12, 6

And.....the first term of this series  = 24 (2)^3  = 192

And the sum of this  infinite series  = [ first term] / [ 1 - r ] =

192 / [ 1 - 1/2]  =  192/ (1/2)  = 192 * 2  = 384   May 29, 2017