1. G.P

If x,y,z be respectively pth, qth and rth terms of a G.P , prove that (q-r)log x +(r-p)log y +(p-q) log z = 0

2. Originally Posted by matsci0000
If x,y,z be respectively pth, qth and rth terms of a G.P , prove that (q-r)log x +(r-p)log y +(p-q) log z = 0
Hi

x,y,z being respectively pth, qth and rth terms of a G.P (ratio k) you can write

$\displaystyle y = x \: k^{q-p}$ and $\displaystyle z = x \: k^{r-p}$

3. Verify

Can anyone check whether the question is correct or not ?

I am getting a different solution.
My working:

" _ " sign has been used to denote subscript.

Given:
a_p=x a_q=y a_r=z

Let 'a' be the first term and R be the common ratio of the G.P.
aR^(p-1)=x aR^(q-1)=y aR^(r-1)=z
logx=log a+(p-1)log R ------------- (1)
logy=log a+(q-1)log R -------------(2)
logz=log a+(r-1)log R ---------------(3)

To prove that :
(q-r)logx+(r-p)log y+(p-q)log z=0
on solving LHS
I obtained 2 log 0
but log 0 is undefined.

4. You should obtain
(q-r)log a+(r-p)log a+(p-q)log a + [(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]log R=0

5. solution to the question on G.P.

Originally Posted by matsci0000
If x,y,z be respectively pth, qth and rth terms of a G.P , prove that (q-r)log x +(r-p)log y +(p-q) log z = 0
nth term of G.P = aR ^ (n-1)
a is the first term
R is the common difference

x = pth term y=qth term z = rth term
x = aR^(p-1) y = aR^(q-1) z = aR^(r-1)
log x = (p-1) log aR
log y = (q-1) log aR
log z = (r-1) log aR

(q-r) log x + (r-p) log y + (p-q) log z

= (q-r) (p-1) log aR + (r-p) (q-1) log aR + (p-q) (r-1) log aR

= log aR [ (q-r)(p-1) + (r-p)(q-1) + (p-q)(r-1) ]
=log aR [ qp - q -rp + r + rq - r -pq + p + pr -p -qr +q ]
= log aR [0]
= 0

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if x y and z be respectivly p q and r term of a gp show tha x^q-r.y^r-p.z^p-q=1

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