2009/012) If the pth, qth, rth and sth terms of an AP are in GP, show that (p-q),(q-r),(r-s), are also in GP

Let 1st term be a+ t and difference be t

We need to prove
(p-q)/(q-r) = (q-r)/(r-s)
So pth term = a + pt
Qth terem= a + qt
Rth term = a + rt
And sth term = a + st

Thery are in GP then

(a+st)/(a+rt) = (a+rt)/(a+qt) = (a+qt)/a+pt) ( common ratio)

Now if (a/b) = (c/d) then both = (a-c)/(b-d)

Using this we get

From (a+st)/(a+rt) = (a+rt)/(a+qt) both = (st-rt)/(rt-qt) = (s-r)/(r-q)

From (a+rt)/(a+qt) = (a+qt)/a+pt) both = (r-q)/(q-p) same as aboth

Hence
(s-r)/(r-q) = (r-q)/(q-p)
Or
(r-s)/(q-r) = (q-r)/(p-q) multiplying numerator and denominator of both sides by -1

Proved