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

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- August 16th 2009, 05:06 AM #1

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- August 16th 2009, 05:19 AM #2

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- August 16th 2009, 05:25 AM #3

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## 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, 11:41 AM #4

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- August 29th 2009, 06:40 PM #5

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## 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