Answer : Given data is, x2 – 3x + p = 0 → (1)

a and b are roots of (1)

So, (x + a)(x + b) = 0 x2 – (a + b)x + ab = 0

So, a + b = 3 and ab = p → (2) Given data is,

x2 – 12x + q = 0 → (3)

c and d are roots of (1) So, (x + c)(x + d) = 0 x2 – (c + d)x + cd = 0

So, c + d = 12 and cd = q → (4) a, b, c, d are in GP.(Given data)

Similarly A, AR, AR2, AR3 also forms a GP, with common ratio R. From (2),From (4), c + d = 12
a + b = 3 A + AR = 3

AR2 + AR3 = 12

AR2 (1 + R) = 12 → (6)

Substituting value of (1 + R) in (6). R = 2

Now, substitute value of R in (5) to get value of A, A = 1

Now, the GP required is A, AR, AR2, and AR3 1, 2, 4, 8…is the required GP.

So,

a = 1, b = 2, c = 4, d = 8

From (2) and (4), ab = p and cd = q

So, p = 2, and q = 32.

So, (q + p): (q – p) = 17: 15.