Any clues ?
If pth, qth, rth, and sth term of an A.P are in G.P then (p-q), (q-r),( r-s) are in ?
My work so far:-
An algebraic approach:-
Since these terms are in G.P, hence
Unable to move further on the above fraction. It's a huge expression and I am unable to express it into lowest terms.
Next, p-q= a+(p-1)d - [a+(q-1)d] = p-q.
Similarly we get q-r=q-r and r-s=r-s.
But, how do we prove that (p-q),(q-r) and (r-s) are in G.P algebraically.
[Request an algebraic solution]
First thing-since we are given an A.P so the first G.P must be defined in terms of the terms of the A.P otherwise the information regarding the A.P is useless. But there is a catch, which I have figured out after much experimentation with the numbers:- If the pth,qth,rth and sth terms of an A.P are in G.P then p,q,r,s themselves are in G.P. (Hence the requirement of the information regarding A.P). I can give you some examples:-
Let the two A.P be
From the 1st A.P: p=1,q=2,r=4,and s=8 i.e pth term=1st term so p=1, qth=2nd term so q=2, rth=4th term so r=4 and so s=8. The question mentions that p-q,q-r,r-s form a progression. We can clearly see that these form a G.P and the question never mentions the pth-qth term or qth-rth term or rth-sth term from a progression.
Similarly, for the 2nd A.P: p=1,q=3,r=9 and s=27 (since 81 is the 27th term of the A.P). So here we can clearly see that
p-q,q-r,r-s are in G.P. You can try this on any A.P.
We can see that the algebra becomes very easy once we find that p,q,r and s are in G.P. We can assume any values for p,q,r and s since p,q,r,s are independent of the values of pth,qth,rth,sth terms and don't have to relate p,q,r,s to the A.P values. I guess this is the reason that the question mentions the information about A.P.
Now, I need one confirmation. I don't know whether there is any such rule that if the pth,qth,rth,sth term of a A.P are in G.P then p,q,r and s are themselves are in G.P. Please inform whether there is such rule oand whether I have made correct conclusions. Please inform soon.
I don't know whether I will be able to confirm that there is a rule. But it is true that p,q,r and s will be in G.P if the pth,qth,rth and sth term of an A.P are in G.P. I have to see what other members say about this ; what is their opinion before making any confirmation. Even I am interested in knowing what the members in this forum say about this.Now, I need one confirmation. I don't know whether there is any such rule that if the pth,qth,rth,sth term of a A.P are in G.P then p,q,r and s are themselves are in G.P. Please inform whether there is such rule oand whether I have made correct conclusions. Please inform soon.