plz help me with this maths problem_series and sequences_geometric progressions?

Question

hi
this is not any homework . i jus' wanna clear ma doubt
1) the pth term is P ; the qth term is Q . show that the nth term is
[ P^n-q/Q^n-p]^1/p-q

thanks again
tutu

Answer 1

The nth term of a GP is ar^(n-1), so the pth term is ar^(p-1)=P and the qth term is ar^(q-1)=Q. Dividing the first by the second we obtain
r^(p-q) = P/Q
i.e. r = (P/Q)^[1/(p-q)]
From this and the expression for P we see that a= P/(P/Q)^[(p-1)/p-q)], so that ar^(n-1)
= {P/(P/Q)^[(p-1)/(p-q)]}x{(P/Q)^[(n-1)/(p-q)] 1
= P/(P/Q)^[(n-p)/(p-q)]
= {P^(p-q)xP^(n-p)/Q^(n-p)}^[1/(p-q)] 2
= {P^(n-q)/Q^(n-p)}^{1/(p-q)]

This would be a lot clearer if I had the means and the ability to write this out without resorting to / and ^, but I hope that you can follow the argument. After submitting this answer I noticed that the equations marked 1 and 2 did not transmit completely. The last part of eq. 1 is (P/Q)^[(n-1)/(p-q)] and the last part of eq. 2 is
^[1(p-q)]

Answer 2

Definition of a geometric series is:

a(n) = a*r^(n-1)

We know that:

pth term is a(p) = a*r^(p-1) = P
qth term is a(q) = a*r^(q-1)= Q

Solve for a:

a = P/[r^(p-1)]

Solve for r:

P/Q = r^(p-q); r = (P/Q)^1/(p-q)

the nth term is

a(n) = ar^(n-1)

Substitute the above formulas for a and r into the above equation and you will get your result.

Answer 3

sdfg