tell me how this happens in maths_geometric progression?

Question

[r^(p-q)] = (p/q)^1/(p-q)

plz tell me how left hand side is equal to right hand side

tanx
tom

tanx 4 pointing out the mistake

it's like this

r^(p-q)= p/q

r= (p/q)^1/p-q

Answer 1

if r^(p-q) = (p/q)

then raise both sides to the power of 1/(p-q); (p ≠ q. if p=q then
1/(p-q) is not defined)

[r^(p-q)] ^ [1/(p-q)] = (p/q) ^ [1/(p-q)]

r ^ [(p-q)/(p-q)] = (p/q) ^ [1/(p-q)]

r ^ 1 = (p/q) ^ [1/(p-q)] or r = (p/q) ^ [1/(p-q)]

THIS IS TRUE ONLY WHEN p ≠ q!

When p=q , r can be anything (except 0) since
if r^(p-q) = p/q and p=q then we get r^0 = 1 and this is true for any r≠0

Answer 2

That doesn't make sense, where does R appear in the right hand side?

Answer 3

If two numbers are given to be in g.p like:
if a and b are in g.p then b/a= r where r is defined as the common ratio.
So, in this case if p and q are in g.p where p is given to be the first term then r=q/p.
Or if q is given to be the first term then r=p/q.
In this case what does " ^ " stands for?

Answer 4

Since there is no "r" on the righthand side, they cannot be equal in the Algebraic sense.

Answer 5

http://en.wikipedia.org/wiki/Geometric_series

Ok so you want to know that
if r^(p-q)= p/q then r= (p/q)^1/p-q

x^n = y => x = y^(1/n); *

let x = r, p-q = n, y=p/q , fill in inn * and done.