# Geometric Progressions Help

1. Apr 28, 2004

### Olly

I am having toruble with my geometric progressions, in that i ahv ebeen given a question where i am given the 7th and 26th terms of a GP. I am required to find the ratio however, which i could do if i had the first term. Usually i can do this as they only give me gps that are one term apart, and i would divide the top by bottom (say Term6 = 3 and term7 = 4) and woudl end up with term1 = 3/4. How can i do this if the terms are as far apart as they are?

Welcoming any responses here

2. Apr 28, 2004

### MathematicalPhysicist

i think you have too many variables such as a1 and n (the number of terms) that are unknown at least one of them are needed to solve for the quotinent.

3. Apr 28, 2004

### arildno

You know that in a geometric progression, the next term's ratio with the previous is a constant; let's call it x; that is GP(n+1)/GP(n)=x.
But then we must have: GP(n+2)/GP(n)=(GP(n+2)/GP(n+1))*GP(n+1)/GP(n)=x^(2).
Did that help?

4. Apr 29, 2004

### Olly

Thanks for the help, ive got it down pat now :) hope im ready for maths test tomorrow