# Evaluate ad-bc if a,b,c,d are the terms of a geometric series?

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We know that all the consecutive terms of a geometric progression have a common ratio. And it is possible to write consecutive terms as the previous term multiplied by the common ratio.

If a, b, c and d are terms of a geometric series, they can be expressed as :

b = ar

c = ar^2

and d= ar^3, where r is the common ratio between the terms.

So a*d - b*c = a*ar^3 - ar*ar^2 = a^2*r^3 - a^2*r^3 = 0

**Therefore ad - bc is equal to zero.**

We assume that the a,b,c and d are the cosecutive terms of a geometric progression.

So aa =a

a2 = ar = b

a3 = ar^2 = c

a4 = ar^3 = d.

Therefore ad-bc = a*ar^3 - ar*ar^2 = a^2r^3-a^2*r^3 = 0.

Threfore ad-bc = 0.

We'll apply the mean theorem of a geometric series:

b^2 = a*c

sqrt b^2 = sqrt a*c

b = sqrt a*c (1)

c^2 = b*d

c = sqrt b*d (2)

We'll multiply bc = sqrt a*b*c*d

But b = a*r, where r is the common ratio.

c = a*r^2

d = a*r^3

a*b*c*d = a*a*r*a*r^2*a*r^3

a*b*c*d = a^4*r^6

sqrt a*b*c*d = sqrt a^4*r^6

sqrt a*b*c*d = a^2*r^3

bc = a^2*r^3 (3)

ad = a*a*r^3 (4)

We'll subtract (4) from (3):

a^2*r^3 - a^2*r^3 = 0

So, the result of the difference is:

**ad - bc = 0, if and only if a,b,c,d are the terms of a geometric series.**