Since a and b are the roots of $x^{2}$ – 3x + p = 0 and c, d are roots of $x^{2}$ – 12x + q = 0 Then, a + b = 3, ab = p, c + d = 12, cd = q Also, a, b, c, d forms a G.P., then if a is first term and r is a common ratio, then b = ar, c = $ar^{2}$, d = $ar^{3}$ a + b = 3 ⇒ a + ar = 3 ⇒ a (1 + r) = 3 ...(i) c + d = 12 ⇒ $ar^{2} + ar^{3}$ = 12 ⇒ $ar^{2}$ (1 + r) = 12 ...(ii) Dividing (ii) by (i), we get, $r^{2}$ = 4 ...(iii) Now, ab = a(ar) = $a^{2}r$ = p ...(iv) cd = $(ar^{2})(ar^{3})$ = $a^{2}r^{5}$ = q ...(v) Dividing (v) by (iv), we get $r^{4}$ = $\frac{p}{q}$ ⇒ $(4)^{2}$ = $\frac{q}{p}$ [using (iii)] ⇒ $\frac{q}{p}$ = $\frac{16}{1}$ Applying componendo and dividendo, we get $\frac{q+p}{q-p}$ = $\frac{17}{15}$ i.e., (q + p) : (q – p) = 17 : 15