Results 1 to 5 of 5

- August 16th 2009, 04:06 AM #1

- Joined
- Jul 2009
- Posts
- 40

- August 16th 2009, 04:19 AM #2

- Joined
- Nov 2008
- From
- France
- Posts
- 1,458

- August 16th 2009, 04:25 AM #3

- Joined
- Jul 2009
- Posts
- 40

## 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.

- August 16th 2009, 10:41 AM #4

- Joined
- Nov 2008
- From
- France
- Posts
- 1,458

- August 29th 2009, 05:40 PM #5

- Joined
- Aug 2009
- Posts
- 2

## solution to the question on G.P.

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

Click on a term to search for related topics.