# Homework Help: Relationship between A.P & G.P

1. Oct 5, 2012

### Anyiam

1. The problem statement, all variables and given/known data

the Pth, Qth & Rth terms of an arithmetic sequence are in geometric progression. Show that the common ratio is (q-r)/(p-q) or (p-q)/(q-r).

2. Relevant equations
for an A.P, the Nth term=
a (n-1)d
for a G.P, the Nth term= ar^(n-1)

3. The attempt at a solution

let the Pth, Qth & Rth term be
p, q & r respectively. Since they are in A.P,
"d"= q-p = r-q
also, since they form a GP,
"r"= (p/q)or(q/p)= (r/q)or(q/r)
don't really know if to make the assumption that for this case, "d"= "r", is pretty safe!
So please what do i do next?
Thanks in anticipation!

2. Oct 5, 2012

### tiny-tim

Hi Anyiam!
uhh?

they'll be a + pd etc

3. Oct 5, 2012

### Anyiam

Oops! That was some silly mistake on my part! The Pth, Qth & Rth term ought to have been: a plus(p-1)d,
a plus(q-1)d, and
a plus(r-1)d respectively!
pls permit me to use "plus" to indicate addition.
Going by this, the difference between the
Qth & Pth term becomes
d(q-p) and between the
Rth & Qth term also becomes d(r-q), & equating both gives:
d(q-p)= d(r-q).
So what do i do next?

4. Oct 5, 2012

### tiny-tim

what are you doing?

the question says …
ie (a + pd)/(a + qd) = (a + qd)/(a + rd)

5. Oct 5, 2012

### Anyiam

Ok! You are right! Because the common ratio must be the ratio between any two consecutive terms say the Pth term[a plus pd] & the Qth term[a plus qd]. But how do i proceed from here?

Last edited: Oct 5, 2012
6. Oct 5, 2012

### tiny-tim

carefully, but with confidence!

show us what you get

7. Oct 5, 2012

### Anyiam

i think the next thing is to eliminate the "d" & "a".

Last edited: Oct 5, 2012
8. Oct 5, 2012

### Anyiam

Well, i eventually did it this way:
[Pth term - Qth term] /
[Qth term - Rth term] and i got (p-q)/(q-r).
and taking the inverse will also give (q-r)/(p-q), which is true!