Question #41113

2 Answers
Jul 30, 2015

Answer:

This series can only be a geometric sequence if #x=1/6#, or to the nearest hundredth #xapprox0.17#.

Explanation:

The general form of a geometric sequence is the following:
#a,ar,ar^2,ar^3,...#
or more formally #(ar^n)_(n=0)^oo#.

Since we have the sequence #x,2x+1,4x+10,...#, we can set #a=x#, so #xr=2x+1# and #xr^2=4x+10#.

Dividing by #x# gives #r=2+1/x# and #r^2=4+10/x#. We can do this division without problems, since if #x=0#, then the sequence would be constantly #0#, but #2x+1=2*0+1=1ne0#. Therefore we know for sure #xne0#.

Since we have #r=2+1/x#, we know
#r^2=(2+1/x)^2=4+4/x+1/x^2#.
Furthermore we found #r^2=4+10/x#, so this gives:
#4+10/x=4+4/x+1/x^2#, rearranging this gives:
#1/x^2-6/x=0#, multiplying by #x^2# gives:
#1-6x=0#, so #6x=1#.
From this we conclude #x=1/6#.

To the nearest hundredth this gives #xapprox0.17#.

Jul 30, 2015

Answer:

As Daan has said, if the sequence is to be geometric, we must have #x=1/6 ~~ 0.17# Here is one way to see that:

Explanation:

In a geometric sequence, the terms have a common ratio.

So, if this sequence is to be geometric, we must have:

#(2x+1)/x = (4x+10)/(2x+1)#

Solving this equation gets us #x = 1/6#