Given: If a, b, c, d are in G.P., prove that (a + b), (b + c), (c + d) are in G.P.

It is given that a and b are the roots of x² − 3x + p = 0.

∴ a + b = 3 and ab = p     ...(1)

Also, c and d are the roots of x² − 12x + q = 0.

∴ c + d = 12 and cd = q     ...(2)

Since a, b, c, d are in G.P., let a = x, b = xr, c = xr², d = xr³.

From (1) and (2):

x + xr = 3 ⇒ x(1 + r) = 3 ⇒ x = 3 / (1 + r)

and xr² + xr³ = 12 ⇒ x r²(1 + r) = 12

Dividing the two equations:

r² = 4 ⇒ r = ±2

When r = 2, x = 3 / 3 = 1

When r = −2, x = 3 / (−1) = −3

Case I: When r = 2 and x = 1,

ab = x²r = 2,    cd = x²r⁵ = 32

q − p = 32 − 2 = 30

(q + p) : (q − p) = 17 : 15

Case II: When r = −2 and x = −3,

ab = x²r = −18,    cd = x²r⁵ = −288

q − p = −288 − (−18) = −270

(q + p) : (q − p) = 17 : 15

Hence, in both cases, (q + p) : (q − p) = 17 : 15.