# The real positive numbers a,b,c,d are the terms of a geometric progression. What is the common ratio if d-a=7 and c-b=2?

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have four terms a, b, c, d in geometric progression with d - a = 7 and c - b = 2. We have to find the common ratio.

Let the common ratio be r, this gives b = ar, c = ar^2 and d = ar^3

ar^3 - a = 7 and ar^2 - ar = 2

ar^2 - ar = 2

=> a = 2/(r^2 - r)

substitute in ar^3 - a = 7

=> (r^3 - 1)*2/(r^2 - r) = 7

=> 2*r^3 - 2 = 7r^2 - 7r

=> 2r^3 - 7r^2 + 7r - 2 = 0

=> 2r^3 - 4r^2 - 3r^2 + 6r + r - 2 = 0

=> 2r^2(r - 2) - 3r(r - 2) + 1(r - 2) = 0

=> (2r^2 - 3r +1 )(r - 2) = 0

=> (2r^2 - 2r - r + 1)(r - 2) = 0

=> (2r(r - 1) - 1(r - 1))(r - 2) = 0

=> (2r - 1)(r - 1)(r - 2) = 0

=> r = 0.5 and r = 1 and r = 2

The possible values of the common ratio are (0.5, 1, 2)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since the common ratio of a geometric progression is the ratio of a term to the previous term, we'll have:

b = a*q

c = b*q = a*q^2

d = c*q = a*q^3

We'll re-write the given constraints:

d - a = a*q^3 - a = 7

a(q^3 - 1) = 7

The difference of cubes returns the product:

q^3 - 1 = (q-1)(q^2 + q + 1)

a(q-1)(q^2 + q + 1) = 7 (1)

c - b = a*q^2 - a*q = 2

a*q*(q - 1) = 2 => a*(q - 1) = 2/q (2)

We'll substitute (2) in (1):

(2/q)*(q^2 + q + 1) = 7

2q^2 + 2q + 2 = 7q

We'll subtract 7q:

2q^2 + 2q + 2 - 7q = 0

2q^2 - 5q + 2 = 0

q1 = [5+sqrt(25 - 16)]/4

q1 = (5 + 3)/4

q1 = 2

q2 = (5 - 3)/4

q2 = 1/2

The values of the common ratio of the geometric progression a,b,c,d are: {1/2 ; 2}.