Question #41113

2 Answers
Jul 30, 2015

This series can only be a geometric sequence if x=1/6x=16, or to the nearest hundredth xapprox0.17x0.17.

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

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