Solve this progression problem?

For the AP, convert all the terms to base a using the change of base rule to get

the progression (1/log(c)), (log(c)/log(b)), log(b), where log is log(base a).

Now log(c) = log(a*r^2) = log(a) + log(r^2) = 1 + 2*log(r), and

log(b) = log(a*r) = log(a) + log(r) = 1 + log(r).

Thus the AP becomes (1/(1 + 2*log(r))), ((1 + 2*log(r))/(1 + log(r))), (1 + log(r)).

So the common difference d equals both (1 + log(r)) – ((1 + 2*log(r))/(1 + log(r)))

and ((1 + 2*log(r))/(1 + log(r))) – (1/(1 + 2*log(r))). Letting log(r) = x this gives us

(1 + x) – ((1 + 2x)/(1 + x)) = ((1 + 2x)/(1 + x)) – (1/(1 + 2x)) —->

2*((1 + 2x)/(1 + x)) – (1/(1 + 2x)) – (1 + x) = 0 —-> which for x not -1 or -1/2 gives

(2 + 4x)*(1 + 2x) – (1 + x) – (1 + x)^2*(1 + 2x) = 0 —–>

2 + 8x + 8*x^2 – 1 – x – (1 + 4x + 5*x^2 + 2*x^3) = 0 —->

3x + 3*x^2 – 2*x^3 = 0 ——> x*(2*x^2 – 3x – 3) = 0.

So either x = log(r) = 0, which gives r = 1 and a = b = c, leading to d = 0,

or 2*x^2 – 3x – 3 = 0. In this case use the quadratic formula to solve for x:

x = (3 +/- sqrt(9 + 4*2*3)) / (2*2) = (1/4)*(3 +/- sqrt(33)).

So either x = log(r) = (1/4)*(3 + sqrt(33)) or x = log(r) = (1/4)*(3 – sqrt(33)).

Now plug these solutions into the equation

d = (1 + x) – ((1 + 2x)/(1 + x)) = (1/(1 + x))*((1 + x)^2 – (1 + 2x)) = x^2 / (1 + x):

with x = (1/4)*(3 + sqrt(33)) we have

d = (1/16)*(3 + sqrt(33))^2 / ((1/4)*(4 + 3 + sqrt(33))) =

(1/4)*(9 + 33 + 6*sqrt(33)) / (7 + sqrt(33)) =

(1/4)*6*(7 + sqrt(33)) / (7 + sqrt(33)) = 6/4 = 3/2.

With x = (1/4)*(3 – sqrt(33)) we have

d = (1/16)*(3 – sqrt(33))^2 / ((1/4)*(4 + 3 – sqrt(33))) =

(1/4)*(9 + 33 – 6*sqrt(33)) / (7 – sqrt(33)) =

(1/4)*6*(7 – sqrt(33)) / (7 – sqrt(33)) = 6/4 = 3/2.

So we can either have the trivial case with d = 0,

or two non-trivial cases with d = 3/2.
– Chosen by Asker