Marriage of Arithmetic and Geometric Sequence

B i.e. 2/3

we can just choose a = 1, b = 2, c = 3 and d = 4

This satisfies both arithmetic and geometric series requirements as per the question..

Very interesting question!

My aproach:

a, b, c, d would equal in the arithmetic sequence:

a,a+x,a+2x, and a+3x

in the geometric sequence

a, b, d

a/b = b/d= y

so b=a*y and d=b*y^2

Then we have to make equations from the two sequences.
a=a
b=a+x=a*y
c=a+2x
d= a+3x=b*y^2

So a+3x= a*y^3

And here is when I need help from a Math Guru! May be by picking numbers/pluging in answers we can solve this!

El Cucu wrote:Very interesting question!

My aproach:

a, b, c, d would equal in the arithmetic sequence:

a,a+x,a+2x, and a+3x

in the geometric sequence

a, b, d

a/b = b/d= y

so b=a*y and d=b*y^2

Then we have to make equations from the two sequences.
a=a
b=a+x=a*y
c=a+2x
d= a+3x=b*y^2

So a+3x= a*y^3

And here is when I need help from a Math Guru! May be by picking numbers/pluging in answers we can solve this!

What's the point? After doing all this you still want to plugin values.. well, in that case, just plug in values like I have shown above... Wouldn't that be more simple?