If a, b, c are the pth, qth and rth terms of a GP, show that

(q – r) log a + (r – p) log b + (p – q) log c = 0.


As per the question, a, b and c are the pth, qth and rth term of GP.


Let us assume the required GP as A, AR, AR2, AR3


Now, the nth term in the GP, an = ARn-1


pth term, ap = ARp-1 = a (1)


qth term, aq = ARq-1 = b (2)


rth term, ar = ARr-1 = c (3)


(i)


(ii)


(iii)


Taking logarithm on both sides of equation (i), (ii) and (iii).


(p – q) log R = log a – log b,


(4)


(q – r) log R = log b – log c


(5)


(r – p) log R = log c – log a


(6)


Now, multiply equation (4) with log c,


(7)


Now, multiply equation (5) with log a,


(8)


Now, multiply equation (6) with log b,


(9)


Now, add equations (7), (8) and (9).



On solving the above equation, we will get,


(p – q) log c + (q – r) log a + (r – p) log b = 0


Hence proved.


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