If pth , qth , rth and sth terms of an ap are in gp, show that (p-q), (q-r), (r-s) are also in gp

Here is what we got and it can be solved:
a_p = a+(p-1)d
same for q,r and s
given : a_q = ma_p and m^2 a_p,,where m is the common ratio
and (a_p -a_q) = (p-q)d
also (a_q -a_r) =(q-r)d
and (a_r -a_s) =(r-s)d
now (a_p - a_q)(a_r - a_s) = (p-q)(r-s)d^2 = (a_p a_r - a_p a_s - a_q a_r + a_q a_s)
we know that (a_p a_r) =(a_q)^2,, a_q a_s = (a_r)^2 ,,
and a_p a_s = a_q a_r
=>we then have a_q^2 -2a_q a_r + a_r^2 = (a_q - a^r)^2
also note that it was equal to (p-q)(r-s)d^2
so ,, (p-q)(r-s) = [ (a_q - a_r)/d]^2 = (q-r)^2
hence they are in Gp
Hope this is good for you