Given that a and b are roots of x² – 3x + p = 0

⇒ a + b = 3 and ab = p ...(i)

It is given that c and d are roots of x² – 12x + q = 0

⇒ c + d = 12 and cd = q...(ii)

Also given that a, b, c, d are in G.P.

Let a, b, c, d be the first four terms of a G.P.

⇒ a = a, b = ar c = ar² d = ar³

Now,

∴a + b = 3

⇒ a + ar = 3

⇒ a(1 + r) = 3…(iii)

c + d = 12

⇒ ar² + ar³ = 12

⇒ ar²(1 + r) = 12.....(iv)

From (iii) and (iv) we get

3.r² = 12

⇒ r² = 4

⇒ r = ±2

Substituting the value of r in (iii) we get a = 1

⇒ b = ar = 2∴

c = ar² = 2² = 4

d = ar³ = 2³ = 8

⇒ ab = p = 2and cd = 4×8 = 32

⇒ q + p = 32 + 2 = 34 and q−p = 32−2 = 30

⇒ q + p:q−p = 34:30 = 17:15

Hence, proved.