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...

Aleguan May 29, 2017

#1**+2 **

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

CPhill May 29, 2017