terms of a geometric series and arithmetic series, find common ratio

[itsgotabigbirdonitdude]

Different numbers x, y and z are the first three terms of a geometric progression with common ratio r, and also the first, second and fourth terms of an arithmetic progression.
a. Find the value of r.
b. Find which term of the arithmetic progression will next be equal to a term of the geometric progression.

I haven't tackled part b. yet but I'm guessing it must be quite straightforward once r is found, but for now I'm having major issues with a.
So far I've come up with r=y/x=z/y (since all are different versions of the common ratio of the geometric progression), z-y = 2(y-x) (since y-x is the common difference of the arithmetic progression and z-y is the difference between the second and fourth terms) and y-x = y/x.
However, I'm confused as to how to combine these equations in order to find r. All of my attempts have turned up hopelessly complex or just plain incorrect. Any suggestions as to the correct and most simple way to go about this would be appreciated, cheers.

 

[Deleted member 4993]

Different numbers x, y and z are the first three terms of a geometric progression with common ratio r, and also the first, second and fourth terms of an arithmetic progression.
a. Find the value of r.
b. Find which term of the arithmetic progression will next be equal to a term of the geometric progression.

I haven't tackled part b. yet but I'm guessing it must be quite straightforward once r is found, but for now I'm having major issues with a.
So far I've come up with r=y/x=z/y (since all are different versions of the common ratio of the geometric progression), z-y = 2(y-x) (since y-x is the common difference of the arithmetic progression and z-y is the difference between the second and fourth terms) and y-x = y/x.
However, I'm confused as to how to combine these equations in order to find r. All of my attempts have turned up hopelessly complex or just plain incorrect. Any suggestions as to the correct and most simple way to go about this would be appreciated, cheers.

By observation, one of the solutions is:

(1, 2, 4) & (1, 2, 3, 4) → d = 1 & r = 2

 

[lookagain]

Different numbers x, y and z are the first three terms of a geometric progression
with common ratio r, and also the first, second and fourth terms of an arithmetic progression.

a. Find the value of r.

Let r = the common ratio and d = the common difference


\(\displaystyle x, y, z\)


\(\displaystyle xr = y, \)

\(\displaystyle xr^2 = z\)



\(\displaystyle x \ (being \ the \ 1st \ no.) + d = y \ \ (being \ the \ 2nd \ no.), \)

\(\displaystyle y + d = \ (being \ the \ 3rd \ no.),\)

\(\displaystyle y + d + d = z \ \ (being \ the \ 4th \ no.) \ \ ------> \ \ y + 2d = z\)




\(\displaystyle x + d = xr\)

\(\displaystyle xr + 2d = xr^2\)


Multiply each side of the first equation by 2 and subtract the second equation:

\(\displaystyle 2x - xr = 2xr - xr^2\)



Divide each side by x:

\(\displaystyle 2 - r = 2r - r^2\)

\(\displaystyle r^2 - 3r + 2 = 0 \)

\(\displaystyle (r - 1)(r - 2) = 0\)

\(\displaystyle r= 1, \ \ r = 2\)



r = 1 gives d = 0, but then x, y, and z would fail to be different numbers.


r = 2 gives d = 1 as Subhotosh Khan typed out

 

[itsgotabigbirdonitdude]

Alright thanks for your help. So I managed to solve a. in the following manner:
r=y/x which means that y=rx, and since r=z/y, z=ry=(r^2)x
also z-y = 2(y-x), and plugging in the above values (in terms of x) of z and y I get x(r^2) - rx = 2rx - 2x
eliminating x from both sides of this equation gives (r^2) - 3r + 2 = 0, which gives me r=1 or 2.

However I'm having some problems with b.
I'm using the equation a(r^(n-1)) = a + (n-1) d, where a=x, r=2 and I need to solve for n.
Other than that, so far I've just been working in circles, and it's pretty frustrating. Where should I start looking for x and d? Do I even need to find x?

 

[lookagain]

Different numbers x, y and z are the first three terms of a geometric progression with common ratio r, a
nd also the first, second and fourth terms of an arithmetic progression.

a. Find the value of r.


> > b. Find which term of the arithmetic progression will next be equal to a term of the geometric progression.

The terms of the geometric sequence are increasing consecutive powers of 2, beginning with 1.

The terms of the arithmetic sequence are increasing consecutive integers, beginning with 1.

The next term in the geometric sequence is 8, and 8 is also a term in the arithmetic sequence.

My answer to the second problem would be 8 as the next term.